# Chapter 1. The Central Limit Theorem

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0:46

### Question 1.1

8z+81cg3KLg3OGJribkkF67Wsf1U2wzNWqPNF/UOVrLf94MbUDaFEIts57pqtdJP5rC1AhwkTaApjOK94qcl37xrmqQ1DNrLkoowaTqjx2pggFyvDToJSda+MzUlMST0xNkRVdj+U+yb6wNwcs7Oy2DRGWKUa22gpguzgwjYctXoGiOf1fCUl6BHkRUxjuh6ETjhthn4CM8sct1bALZG+Nphc/RqoautgzsrCPaX9g33f1jjO6IY0A7wrwyyxa/GTyNF8ETVPsBnv55zbfnSmw9xeDNOi0v/TvTAg91pPhAm5DKR5udfs8LbkGMqzJleKtEqSN2WBUxOi9e4gncj2FPsfM2bKkl1E425JyR5zR9NieBATU/qxS++wLPkpXMAVLdzQBYSvd2UYrG8fWxtgj423LZ538bmS/Uxc56J3h0Cis0qDTkvAQ==
Correct. A distribution of a variable gives the possible values of the variable together with how often each value occurs. So, the sampling distribution of $$\overline{x}$$ gives the possible values of $$\overline{x}$$ together with how often each value occurs.
Incorrect. A distribution of a variable gives the possible values of the variable together with how often each value occurs. So, the sampling distribution of $$\overline{x}$$ gives the possible values of $$\overline{x}$$ together with how often each value occurs.
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1:05

### Question 1.2

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Correct. The mean of the sampling distribution of $$\overline{x}$$ always exactly equals μ, so the mean equals µ = 40.
Incorrect. The mean of the sampling distribution of $$\overline{x}$$ always exactly equals μ, so the mean equals µ = 40.
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1:17

### Question 1.3

75EfR8DuXrECmYQZjt15p4bORsJHge1oXp9Up33/vHRNsqrQjU/oa0xaO6RtyqwnKkqQZhD9Whq4FcSHmyefPY106IIXSXmXHHOK74MZjT56zZULyQEkevINcG9HUwCV4kLGfX2DhdV9SLUxh7+tWpOSOp4EilkqGKpaXi8uxSrYqJxkvaFcKOiqwWsWWwOe1nZoYRLZplYgpYCxyu1viE0A8B/UE+wGpyMz9jZsY1zGjaLVCrTqTuCbcPq6huV8Lx5jN3jLiX1/Pk0f5/eQOZJBIfh0Lm5YIlPCA2FdztdkRGXrDKOlIZGSo2ihH0gaRQqSx7Gd5PJhoVywI8uzz7XH85tf8BoWewVrrf+YZutQQSDPk6p0Cima+B4=
Correct. The standard deviation of the sampling distribution of equals $$\frac{ \sigma }{ \sqrt{n} }$$ regardless of sample size and shape of the population, so $$\frac{ \sigma }{ \sqrt{n} }$$ = $$\frac{ 12 }{ \sqrt{9} }$$ = 4.0.
Incorrect. The standard deviation of the sampling distribution of equals $$\frac{ \sigma }{ \sqrt{n} }$$ regardless of sample size and shape of the population, so $$\frac{ \sigma }{ \sqrt{n} }$$ = $$\frac{ 12 }{ \sqrt{9} }$$ = 4.0.
2
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1:51

### Question 1.4

Yi6ekr/imq17D7Ukc8EOcobvOgW91DtHq4TvmIClYn0OP+NQbMGx7y96D1MjbVb46hQtRazJNQEzrY0FxgBlQiksdFgJUe8NibWud6o9KJ6Odwdvn1qxlrkSrG9LYdraxH5ejsW0KL6mb589EEbTTrGO2njp4T42NtrrxjIi3BbD/6+yf0AYnY4DgJrQubnyFGcv3ZM7FsRB5mHXB03pvM7VjpmfcCV283FGseB2b7CWLEGSQzULP1hvZVZiLjE68a98eT3iyDaqNUFY/rFPATybuc83BetoBs2I7dnKxu+52YZOSb01LKbO7jijQzZuPMRGVt8TE73YY21z8HJSrLRuyf1aZshkHDbyzHjcxMQpsaaB9MkPSYDhDMcsjjhmxu0rckb/el3MW0dFuaI2fqGAxG2pVQnh8bY7ZZFc2cg5CL9GzM4E1SOJeim2enDqzCXsbyAI+i1lvj6p
Correct. For the sampling distribution to be Normal, the population shape must be Normal; for the sampling distribution of $$\overline{x}$$ to be approximately Normal, the sample size must be large. Neither of those conditions are met in the description of the problem.
Incorrect. For the sampling distribution to be Normal, the population shape must be Normal; for the sampling distribution of $$\overline{x}$$ to be approximately Normal, the sample size must be large. Neither of those conditions are met in the description of the problem.
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### Question 1.5

5VKqnsDebJgKdRmOYKcep07H32KQ+i0m618QNXr+CciMLDeHPZT1ut9Pcb+DTmXhrxXAX/GEBzOyY7ufZIIw4jK30uVQw8YBKMqI58szuZwIoWTw7O/KcI451YNPPH9B/nrJazwH14kXSFjwKtfM9sT33Yxoo9Bbdo1jc+a3FJq5ewqS7KzJIc5eDuLQ4xV99XC3kEl4wANRzwjdZXC8JvDXgzkK38jguSKVx376099/+c9jwcV8zEMks23J5fePtL3ZcOpxzNUIsRy64qsb+36hvUC6CP+B5JsuDAhBOglP0kzPY76pzSjcOeAiGGc5JC06fABsB46rlIDjHuL3ykjFsBszgydKrzn5VbY9zrcEXlzUNpJUNA0EDNcBGrODKpRvRHBw9cw=
Correct. For the sampling distribution to be Normal, the population shape must be Normal; for the sampling distribution of $$\overline{x}$$ to be approximately Normal, the sample size must be large. The population is described to have a Normal shape.
Incorrect. For the sampling distribution to be Normal, the population shape must be Normal; for the sampling distribution of $$\overline{x}$$ to be approximately Normal, the sample size must be large. The population is described to have a Normal shape.
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4:42

### Question 1.6

W5qNaKpboL1fX/hrCOGn5wnIT+2rwJ/P9fnPQ6H2ryuJr1/OS2WHL49r1OpAAimbCBbhv7a3PyLNVbzeRWGjjVkayD2nzTLSVBA8sljKNQM69l7782UH+cLKfZaqe9fIhMg4oMmu5uiLV6tYjAKHQgGOJIABh7ktWrsyEeiNH0IFTHxbcxVf8JpObk6fixo/W6RIkYLmv1EsF4BXc4PbrBTlGC/vEHu5Eu6motVdBh3QB8vAaooL4P2WEpTVQsO3ZLcMIiJA1HLcKsakk3eBZeW4cbgGhbCglGEpnZI7MpbaFxSOvQ6zPA==
Correct. The Central Limit Theorem says that the shape of the sampling distribution of $$\overline{x}$$ is approximately Normal if a large random sample is taken. So the Central Limit theorem has to do with the shape of the sampling distribution of $$\overline{x}$$, not the sample and not the population.
Incorrect. The Central Limit Theorem says that the shape of the sampling distribution of $$\overline{x}$$ is approximately Normal if a large random sample is taken. So the Central Limit theorem has to do with the shape of the sampling distribution of $$\overline{x}$$, not the sample and not the population.
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### Question 1.7

IVH52NZl3Ime2OrSRl/IWeGCeI2oVdiSLK7hJjO+1r12EOcBD1sImJhu3BHn2yhMjBhW7xcRURTSRK5X2k6eXCiBe/LxGsVZT5ZRjWJOppf/KznQYE9dt1lW4OYFISW6KiOXJ714SzXvIt8CQ2IzNcwegWJ9V/g1hH/QoE+7sMIX7y5t0BkCRGnktiYpM3BRFvTQ6fQWSjvn+EneKcN9sAIvcyPjxPJ9uN74tIlUBh0A14KMs9DUaefZlBAwPhIf6xkJzUSt24slU0Ui8ZLOOhIgZecMX6zg3TvT+BJoyiA905ZJ2Yu/5H/zVYsQNon74q/zjzfKlHyzyQhwub9o0xN04FhHG7FbGlAHsoWxkY+dTssK2/e+pv/ET4penU74gAYHMsvrQVwtC/kKyzWiNt3uup/BVxJUy+t3NiqeHdKFSGd2VmkiIiHacHHgRqEg
Correct. Central Limit Theorem says that the shape of the sampling distribution of $$\overline{x}$$ is approximately Normal provided the sample size is large and SRS when sampling from a population having any shape. When the distribution of all the $$\overline{x}$$’s is approximately Normal, we can use a Normal curve to compute a probability on $$\overline{x}$$.
Incorrect. Central Limit Theorem says that the shape of the sampling distribution of $$\overline{x}$$ is approximately Normal provided the sample size is large and SRS when sampling from a population having any shape. When the distribution of all the $$\overline{x}$$’s is approximately Normal, we can use a Normal curve to compute a probability on $$\overline{x}$$.
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### Question 1.8

CY7CTs9aTt1opBWv535rCDaeGRYKtRWFZEieg+7UcNEQxr0K7pQbg8v0QMVuWn5UCPHVlx8C1H2x29vw02/mYvXHawxGuR0f2zTroqKOMnN4wSOJNVUe+uyJgUSTCbEzLixgcMssNuL8wgfCMiJ2SWvh0qze9zl9NPdd+Sr23qMvqfdzNHs+0DpubDU=
Correct. n > 30 is considered large enough to apply the Central Limit Theorem although there is no theoretical basis for this number.
Incorrect. n > 30 is considered large enough to apply the Central Limit Theorem although there is no theoretical basis for this number.
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### Question 1.9

o4PEAWZTjhV4F4uV41SbKfefi16+pTvHWfCUVEIjC2C+R6+W3hqIkh3xsqBM7TjB7hiwCnrvqdGoE7ARty2OzuTRL6SYuQRrF3skcL7XG3LpXTYtJdJ2YPiSJeMgCyaFaG1qBR2So9EvMS5j0s1dz87Mf/zoWuwpS1a9qqJ4Qv1qUZERGZ1TJaPu3MoOME7huhBqpZ/fr55PKIGYYIf8UFshGkvC/ezapCKxUTC/9XdYIaPpaj++G0k+dvHNILFNWXCixiSZMJo=
Correct. Actually the opposite is true—the shape of the sampling distribution of $$\overline{x}$$ is closer to Normal for larger sample sizes than for smaller sample sizes.
Incorrect. Actually the opposite is true—the shape of the sampling distribution of $$\overline{x}$$ is closer to Normal for larger sample sizes than for smaller sample sizes.
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5:02

### Question 1.10

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Correct. The mean of the sampling distribution of $$\overline{x}$$ equals µ for all sample sizes.
Incorrect. The mean of the sampling distribution of $$\overline{x}$$ equals µ for all sample sizes.
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### Question 1.11

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Correct. Facts about the mean and the standard deviation of the sampling distribution of $$\overline{x}$$ are not part of the Central Limit Theorem and are valid regardless of sample size or shape of the population. The facts do require a simple random sample.
Incorrect. Facts about the mean and the standard deviation of the sampling distribution of $$\overline{x}$$ are not part of the Central Limit Theorem and are valid regardless of sample size or shape of the population. The facts do require a simple random sample.
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7:43

### Question 1.12

/ClWhlp9V/V6Cz35buNdWspS43b1CJniWpzjjeMmBE6UlROB/96TwOpw5m6pCkiyoKkgnsLvJlo9y20kZ1GCY/B45/vRVCz3wQcEluLzXrgkV5UH91GcQaAjs3GdQvqBGKFS8loMoV3CFykcXghlGGvmrGJYg8gF1d+QT58erNCVBJXm
Correct. This is a true statement.
Incorrect. This is a true statement.
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### Question 1.13

h/8eLy3/2ImNw1w9oKSpEPWJMm2lqJwyzCu7pfdAQKjIwLJKm3A9yHn1mdAzwzdia1wFoDv/JBAUSv7ObjuCQdiN3hRNj7j6AIORwoCls3puYnIWHSlSVutsQfVH6EoAbZS6ZpfIEjya6TyesxQqYSPog/zS/uPs9WFzugbenqjBokD80ljnydaOnJOTSBiv0JrXgA==
Correct. The standard deviation of the sampling distribution of $$\overline{x}$$ equals $$\frac{ \sigma }{ \sqrt{n} }$$ (not σ) regardless of sample size.
Incorrect. The standard deviation of the sampling distribution of $$\overline{x}$$ equals $$\frac{ \sigma }{ \sqrt{n} }$$ (not σ) regardless of sample size.
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### Question 1.14

Correct. This is a true statement.
Incorrect. This is a true statement.
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### Question 1.15

6DQPrGgGSiD9CA7dbrOhEsNjp8/7CNk/sypFcrQMQ5pFhPQoRuv/jCQlO1BAlHnUVoM47Rp+vdcMZIL+k7MuWkZ5uWGdqBTr1YTENoAcf8BQVAHsvNJxCyuievpvher1lvhXdiVV3y3MBHRRdd3RsnA4UWfRr2cYvQlo4q/Rk0vEaWD8AR24JBhMxrLjNYitQI16Xtc8o+/UYynp72Ju+MtOMfISRRCzXGsAt2y3zN9jHef9
Correct. When sampling from a non-Normal population, the shape of the sampling distribution of $$\overline{x}$$ (not the shape of the histogram of the data in the sample) gets closer to Normal as sample size increases.
Incorrect. When sampling from a non-Normal population, the shape of the sampling distribution of $$\overline{x}$$ (not the shape of the histogram of the data in the sample) gets closer to Normal as sample size increases.
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8:50

### Question 1.16

e/MUSKp48GIExJFu5eKpUEkCSOrTKzM85hji0VRAMLJ6CsmsudelQTdIJN2e7TsCoaBDttzURxKNzT8RuxrG3m1se8wc68aPAiT3nJA4dCSOEp+xMLikMAoSnm2MgvVhwKIUAWuosEAYmmM7UwdL75SQb+Pe0vCmU8tgRYeSWVhY/5zH3a5BKkXI4FqyGY2FyKHRoACS+fTyczwyPXISb75Gj7X5TtJWFHTYF0+wajGIYJxrSz7UFvQXcGkgTNtAPw9qj2RCiy6PMWxe7EEbX3HIEbixz52CqY4lFeyCQVLt1hjn/Pcmn8tN3FOBuBgtLm50CM3k3a10maqoTidEVMbMPEpQYzugLlkvYyu0KamTslFjAkmlL8yQsA7CW7qa4c74MxrGHy7oFjd62BkTJtvErBOMS1/V9SbYyA==
Correct. When finding a probability on $$\overline{x}$$, we must use the standard deviation of the sampling distribution of $$\overline{x}$$ (namely $$\frac{ \sigma }{ \sqrt{n} }$$) in the denominator. Thus, the appropriate z-score is $$z = \frac{ \overline{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } }$$.
Incorrect. When finding a probability on $$\overline{x}$$, we must use the standard deviation of the sampling distribution of $$\overline{x}$$ (namely $$\frac{ \sigma }{ \sqrt{n} }$$) in the denominator. Thus, the appropriate z-score is $$z = \frac{ \overline{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } }$$.
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10:12

### Question 1.17

Z05X92XBc96TT69KByqEeGTUGvSfdLcNAEvEuH+Pt7SQlNnURO5hOr2oJgdpQhzI861PUSN1s/bTk7sWc2aMwCpoaYR5GUalkk76D/kpUkQZHsHKkG+l3iZ5vbXM4n7XBwFNoSg+V/HI9jZC/us9Qs8BVTl7VPp8mOxFz9We9yhKQHvIcNkqLRnGKM2wkkPGgFI0S4q/wB3BJPcl0lQiLg0+wto6Z3fzkZpA1toMpFYpNJeXcwuilMV9xP6w6M5FgpRO+IVsynKzs7MXr1P9cMTcZFm04G8IEm+4jLIdEGbIij7EurWeqhdRy/FWqSY/GlDtt1k6iOY=
Correct. When finding a probability on an individual x, we must use the standard deviation of population (namely, σ) in the denominator. Thus, the appropriate z-score is $$z = \frac{x - \mu }{ \sigma }$$.
Incorrect. When finding a probability on an individual x, we must use the standard deviation of population (namely, σ) in the denominator. Thus, the appropriate z-score is $$z = \frac{x - \mu }{ \sigma }$$.
2
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11:38

### Question 1.18

4J7ZNGE1qfAtMv/sBj1l2cC8Df5hAZR6JRO3ZDDb409H9hVUbElOaCO5sLHDIGmkApwEyJL1LD4RY7DW3fzrjCMkA3EGlcG/Cprkh9snsaa6G0SgIBV32Fs+TfgwBlDIdMaoLNvilMTyYK1vvmpXW2ETeUx/kMZz9B6wQs8fx/bvzOnoetQqxhWl8tC7KZOdYCJ4mcpj+gMSNjDn57L+Gjrx4DOB8vrgOClbwSOmQbKsK3AenKNo/cDwosG7b+bymJN5U3uMQSL0XOalN2fiA2UMxy8NLYHUX/TStiEGJ2GlFxSinAwuIwhAUGuWOnLA4fcYWdFvgOeBcdTHazkgsM7bZNR05/c2WBpD9FzNKxhlrHrzqNh+FPR7qc+TeBi9lD0EUiCTrQbrYegzH5RCksdhV9qw7q77
Correct. The standard Normal table gives area on the left or “less than” probabilities. We needed to find the probability that a randomly selected bottle weighs “more than” 1.1 pounds which is area on the right.
Incorrect. The standard Normal table gives area on the left or “less than” probabilities. We needed to find the probability that a randomly selected bottle weighs “more than” 1.1 pounds which is area on the right.
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### Question 1.19

Correct. Since the population distribution is Normal, we could use the standard Normal table to find the probability on the weight of a randomly selected bottle.
Incorrect. Since the population distribution is Normal, we could use the standard Normal table to find the probability on the weight of a randomly selected bottle.
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12:56

### Question 1.20

Correct. When finding a probability on $$\overline{x}$$, we must use the standard deviation of the sampling distribution of $$\overline{x}$$ (namely $$\frac{ \sigma }{ \sqrt{n} }$$) in the denominator. Thus, the appropriate z-score is $$z = \frac{ \overline{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } }$$.
Incorrect. When finding a probability on $$\overline{x}$$, we must use the standard deviation of the sampling distribution of $$\overline{x}$$ (namely $$\frac{ \sigma }{ \sqrt{n} }$$) in the denominator. Thus, the appropriate z-score is $$z = \frac{ \overline{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } }$$.
2
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14:56

### Question 1.21

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
Correct. Since the population distribution is Normal, the sampling distribution of $$\overline{x}$$ is also Normal for all sample sizes. Thus, we could use the standard Normal table to find the probability on the mean weight from a random sample of eight bottles.
Incorrect. Since the population distribution is Normal, the sampling distribution of $$\overline{x}$$ is also Normal for all sample sizes. Thus, we could use the standard Normal table to find the probability on the mean weight from a random sample of eight bottles.
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15:29

### Question 1.22

Correct. Whenever the population distribution is Normal, we can compute a probability on an individual x and a probability on a sample mean using the standard Normal table.
Incorrect. Whenever the population distribution is Normal, we can compute a probability on an individual x and a probability on a sample mean using the standard Normal table.
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16:10

### Question 1.23

Correct. We can only compute a probability on an individual when the population distribution is Normal. The population distribution of closing stock price is right skewed, not Normal so we cannot use a Normal distribution to find a probability on the closing price of an individual stock.
Incorrect. We can only compute a probability on an individual when the population distribution is Normal. The population distribution of closing stock price is right skewed, not Normal so we cannot use a Normal distribution to find a probability on the closing price of an individual stock.
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17:42

### Question 1.24

L7ifEyVUP6jiaVBN8L9hgvmtbgq8nHEkCPe+8UGRCXvTG1DC/aUIPjRt9pn0xFTBxv4hILCqrYOfj+h2LiYae6LPq4dOCNN7nEP3+rv8MYXVrxYJF7k9sxSY4TwiqwtzEIVUiRVqON8oLXXyF6cqvNzc66HZK2NDMGy3g30Z2v5o4IgjHn4QX9ea9w8bLiye5pZ46tFLzSVO1/o+Jk+W9YvNJO8yk3A8EaVC8dB8E83FDLez5FRy5ooOaqUyM29BJDOJk3mE676P65S53aHytN7R0Y8aNjOea//7Dq6yc0iLkQIXJdaT+1duukrdkD450rQzKnrgUHs9gO35xNgYFcq6tdw50bI+NHe6rNUSgNeQ+RwtWL10bWOVcsYhKbLGOyMsz90l9HX2A/y19nPHYnNfra0=
Correct. Since the sample size is 32 and the sample is random, we can apply the Central Limit Theorem and say that the sampling distribution of $$\overline{x}$$ is approximately Normal.
Incorrect. Since the sample size is 32 and the sample is random, we can apply the Central Limit Theorem and say that the sampling distribution of $$\overline{x}$$ is approximately Normal.
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18:35

### Question 1.25

Correct. When finding a probability on $$\overline{x}$$ (sample mean), we must use the standard deviation of the sampling distribution of $$\overline{x}$$ (namely $$\frac{ \sigma }{ \sqrt{n} }$$) in the denominator. Thus, the appropriate z-score is $$z = \frac{ \overline{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } }$$.
Incorrect. When finding a probability on $$\overline{x}$$ (sample mean), we must use the standard deviation of the sampling distribution of $$\overline{x}$$ (namely $$\frac{ \sigma }{ \sqrt{n} }$$) in the denominator. Thus, the appropriate z-score is $$z = \frac{ \overline{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } }$$.
2
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19:35

### Question 1.26

iBnF5nWkHTkKw820OKDq/0dSB2MiW/L59Wtx9pLqvxh21Sb6vqEmZvvqOEKd0aqa7kgYNdNqFTJYAifmY0LNaOhRFDN4RZHYPXiHJCbYrKU56QaiLHOY3j+U34dX3DvsQl32/L95421rRQvXJH465FgPthZzjeb3H2KKzsO87jRd5qNY1abaOOaTa74QVopes15FHyIltLdn7mt96Go99AgDbkGlYM4ky83ol9p9t/dLVA2a0vTDSDFBzj6dB9Lj8RnT4N57iKkNEoJwpCZBW5ZwOIZqPwXHUYHog/vD6dyCJFKTazD2LaLIMKyWh4A5hgbaQh5HFL7CsgdaGsqnaxYnrWk5of4H
Correct. The standard Normal table gives area on the left or “less than” probabilities. Since we want a “less than” probability, the number found by looking up z = –3.11 is the correct answer.
Incorrect. The standard Normal table gives area on the left or “less than” probabilities. Since we want a “less than” probability, the number found by looking up z = –3.11 is the correct answer.
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### Question 1.27

Correct. In this case, we apply the Central Limit theorem. Even though the population is right skewed (non-Normal), the sample is large and random so the sampling distribution of is approximately Normal allowing us to use $$z = \frac{ \overline{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } }$$ and the standard Normal table to find the probability on the sample mean.
Incorrect. In this case, we apply the Central Limit theorem. Even though the population is right skewed (non-Normal), the sample is large and random so the sampling distribution of is approximately Normal allowing us to use $$z = \frac{ \overline{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } }$$ and the standard Normal table to find the probability on the sample mean.
2
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21:04

### Question 1.28

XWA/aBzoO7X1owDMw2xIAqxPNtplsFHk+RhqChyZb/D0R0inftZexo9ws9vHKkDLCTpE/Z7L96FSlKEiNgPd+0CW67AmkKU5CaeJC7S6U92Z9A4VQllto3GzpFtmoblyH0dDs3iIvaKcQPMsc49dW7wj5+4Nyjec61JBYpi+mzhtpNU/CiIjP7xLjHOdRTzcmL18rcloJVQLs2xgtfYD3tl+dklg3H++DxqjJl9yjH9R23wyGdQaY4hXIQYCEZ86U1uwHQ7vKEIICgfQ1a6rQ2mg3cqGn77H+BoYBK2ujKE9cHSV9lD2rE6GQiqIF+ZSkOzHAt71xxh8vNQq3zK8+v/T4WCBWA5jXf79ZSDg5JJ+erQcdICDXL3DTE9rMEfu9+jmFs08I1P8tkFWed3Y7LBvwjs=
Correct. This is a true statement and will be used repeatedly when we do inference.
Incorrect. This is a true statement and will be used repeatedly when we do inference.
2
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22:17

### Question 1.29

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
Correct. While the shape of a sampling distribution of $$\overline{x}$$ is not always Normal, it is when either the population shape is Normal or the sample size is large. Further, the mean of the sampling distribution of $$\overline{x}$$ always equals µ and the standard deviation of the sampling distribution of $$\overline{x}$$ equals $$\frac{ \sigma }{ \sqrt{n} }$$. These facts allow us to compute a probability on $$\overline{x}$$ without creating the sampling distribution of $$\overline{x}$$.
Incorrect. While the shape of a sampling distribution of $$\overline{x}$$ is not always Normal, it is when either the population shape is Normal or the sample size is large. Further, the mean of the sampling distribution of $$\overline{x}$$ always equals µ and the standard deviation of the sampling distribution of $$\overline{x}$$ equals $$\frac{ \sigma }{ \sqrt{n} }$$. These facts allow us to compute a probability on $$\overline{x}$$ without creating the sampling distribution of $$\overline{x}$$.
2
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24:44

### Question 1.30

Correct. The population consists of all stocks so the histogram of all closing stock prices is a histogram of the population.
Incorrect. The population consists of all stocks so the histogram of all closing stock prices is a histogram of the population.
2
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### Question 1.31

m9x+k8eUpIuXoYmW1FgGwRIaGF0h74HsPNP23hgC+h+MpY1LxEWcqReK4flbh2gi3MDXM/Xe44sXvKFWmk4W8iARp2my8e4vSyB0AwqMN6w8gzE+bOBlrhhJ6WNk/FrtrEfu1/c4Opkav+dAlpNOHDc9PX+H6L2LAb2l4F7lhoENALv0H8yT/T7TR6rQpYONVq1C8RWBZCaGGDlpsmjBIbVnGmez6KgrnwQ8krFmRu8irisKsUlgFoQZ3F3VwJYFgYSwsX7nFRXy5tesOYu/p2UMeOh0bM6Lray1xd69vhBxsWaMsXL4IGhluiqfqzAoVkkGs2HA+I11RtMbmVv+zJ2rD9f0hIn5ZoqjExGSmaA=
Correct. The sampling distribution of $$\overline{x}$$ gives the possible values of $$\overline{x}$$ together with how often each occurs. So the histogram of $$\overline{x}$$’s displays the sampling distribution of $$\overline{x}$$.
Incorrect. The sampling distribution of $$\overline{x}$$ gives the possible values of $$\overline{x}$$ together with how often each occurs. So the histogram of $$\overline{x}$$’s displays the sampling distribution of $$\overline{x}$$.
2
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### Question 1.32

XmMuCLNkb6icmpFvLExKmqx7rkmNFpH3AwVVghVchdTt0dNhyFYtEJU8NIs7AL13r75C1/BoeZZ7Yfgk+AGNm5+1ep12oguXijH1KEBRTfx4+R5U3ENTIRgNMPWTD78ndttyrKHYD8U1Zbpq8wQlRl9Ih9zpejSIAvo8jfNXGWez4xuYvFIwSEOVtTCl0+Jj0iYiaoDRcXHd37FIfgktv8DqwsKemTt0SNE4PJEfHQna1EQlXEALKMReftSXQDcVq6s5P6wGbfdDesF4iSFN3tHiLzbT8YakJ1QbNEU85A+6+9tVwxLO0MDDyfczM7Gk7PLbpPf4SfdLmoPgLjvML7oduh/ejK970JQyF+byf6ZPfheO
Correct. The histogram given in purple in the middle is a histogram of a sample of closing stock prices.
Incorrect. The histogram given in purple in the middle is a histogram of a sample of closing stock prices.
2
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26:02

### Question 1.33

ZEuSnLhklWmQOgqksgYm6dLt5UbfV9DIGqf9Hly0KHmvIJfhNvnXL78M+Cu3+QNELM8y2wRq/Jju6t22mRHN6o/5Ylcm5uR9u9jsTBD24AaCDUkoMSjHaLL9Oe1OATpMBljW6b3hDR9UIwpqRVTbG7d4CuG+cIdWRaR2wg==
Correct. μ is the symbol for the mean of a population.
Incorrect. μ is the symbol for the mean of a population.
2
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### Question 1.34

OaL/Gj2LoR6aMStTRQcaPLA+qX4D/0vT19I9l3jm7lL1MlD/c5gTpaiZejybuQqI1Fb3a71PuIohmvF9aAFH+vL1coHWTIoEOWvfyRW/rpAFLLdEwpQD6ddc8eC+iox+/CGhUVvgaxRDTdYnah4L/TRgkswbr7ZecI4U7SfeF7w=
Correct. $$\overline{x}$$ is the symbol for the mean of a sample.
Incorrect. $$\overline{x}$$ is the symbol for the mean of a sample.
2
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### Question 1.35

Correct. μ is the symbol for the mean of the sampling distribution of $$\overline{x}$$.
Incorrect. μ is the symbol for the mean of the sampling distribution of $$\overline{x}$$.
2
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26:17

### Question 1.36

Correct. σ represents the standard deviation of a population.
Incorrect. σ represents the standard deviation of a population.
2
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### Question 1.37

kLpKWlm+Nn0BZtLZwQmXWCBaxHOHHTTcsWWamYT2WxNMZTJRL5F4EB1aTRPp2VM/75kxGX9wn04AF4HViNRB+E0JDBgNQKtxEcsfPqksPLTWYIvkYMzwvB7kfH64EPowr0sqLk1KM1XxClFQrggDeaUsVcdG7W1/IQh3+p5FG9EBJddnb56EPl/wrk73OV17
Correct. s is the symbol for the standard deviation of a sample.
Incorrect. s is the symbol for the standard deviation of a sample.
2
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### Question 1.38

rkuJPu56fnTYvCbFzsV/zvxCHykpFy1ZqvWbWDmY/pfoeZabAozR1xw60oyNcuU/LrinXBeOaPQEgTwsAO07Z2HwIgNk+uV72v1KbSoaBw4Xlgq8Ew3RCbEWuXeVyDYBOQOtYTTuGF5LIeR+TjAUYVloJquxnN3c7KQO6rXVNecgXrj/Lrzk35XG9oU1qswgDFh04IiDLGBKAe4H2d3q55q2DiJ3+xt6mnISqMatiGIJxQGc
Correct. $$\frac{ \sigma }{ \sqrt{n} }$$ is the symbol for the standard deviation of the sampling distribution of $$\overline{x}$$.
Incorrect. $$\frac{ \sigma }{ \sqrt{n} }$$ is the symbol for the standard deviation of the sampling distribution of $$\overline{x}$$.
2
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27:14

### Question 1.39

+rrsM9BO4T667QuSGrZTbEqKibC4pdht4Eo/qCEGLQyxaf+8QsrI71kJRiuNB0NTl9AXZVVQFKaPzdpFF+w3aV9iJOP3Pa6IKCZdCRq/ohSuxCBnVvtO0lt5gXgXfcXIkygtGNmID3wc4nF8NCPcxsnz47FW2JvqJBzZr0CbQa+Pls3wl3T8BNeYM/oQMIp1h3ckAMD5pYnDryz4rrmmM4166hVnb4lZ2jdGx603imwNMAJDnS1HStpc/r0K8eNY15lDL97LRWekRmcDaRowliwLcUDFsqBrCkOxDNPIlemQeZMbSnKXNqJTJ7wiAt/dpV0AnnjBARZ7fGAUggkHXFtfJNI3zslMl0p0ag==
Correct. If the sample size is large, the shape of the histogram of data in a sample will be approximately Normal and gets more Normal as sample size increases. If sample size is small, we cannot predict the shape.
Incorrect. If the sample size is large, the shape of the histogram of data in a sample will be approximately Normal and gets more Normal as sample size increases. If sample size is small, we cannot predict the shape.
2
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### Question 1.40

Correct. If the sample size is large, the shape of the histogram of data in a sample will be approximately the same as the population distribution and gets closer to the population distribution as sample size increases. If sample size is small, we cannot predict the shape.
Incorrect. If the sample size is large, the shape of the histogram of data in a sample will be approximately the same as the population distribution and gets closer to the population distribution as sample size increases. If sample size is small, we cannot predict the shape.
2
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### Question 1.41

Eu/f04R7OlhUbLS5Mnw7hj9vrY9ULfGC+ND9z9fjbq5pEOSYkZJ26RZYQYOx/OIR8A+ysTCOVcH5H1fCX4mqA6D1Rg6kU8p63rNuF38cSajPbi6hEGJUmCMnYvqxU298mOhTqy2D9Gvt6Y1cFeEuAPfpbbrYlzgkTu7hmxiFNmJ0myh6QYQ9GHKtUYWI9HsTQ+IMlgICosuaHulyfmWMT3oKcacDPJqItBpfbpEiPLaHCclhQ7hvMEhFAINCIi7dwimy5oIsvFRb1p8N2bWwwm5U99hZ9X57zN1DkZCU8AzOUPuS9gzBJts8EW41/VJQDsQfaGW0gR3xFfNvR2FffJV8QyOI0VgV50Zfo6AMpX6RNqS2d20feg==
Correct. If the shape of the population distribution is Normal, the shape of the sampling distribution of $$\overline{x}$$ will always be Normal.
Incorrect. If the shape of the population distribution is Normal, the shape of the sampling distribution of $$\overline{x}$$ will always be Normal.
2
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### Question 1.42

Correct. If the shape of the population distribution is non-Normal, the shape of the sampling distribution of $$\overline{x}$$ will be approximately Normal if the sample size is large. If n is small and the population distribution is skewed, the shape of the sampling distribution of $$\overline{x}$$ will be slightly skewed. There are too many shapes for the population distribution to list all possibilities of shapes of the sampling distribution when n is small.
Incorrect. If the shape of the population distribution is non-Normal, the shape of the sampling distribution of $$\overline{x}$$ will be approximately Normal if the sample size is large. If n is small and the population distribution is skewed, the shape of the sampling distribution of $$\overline{x}$$ will be slightly skewed. There are too many shapes for the population distribution to list all possibilities of shapes of the sampling distribution when n is small.
2
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28:06

### Question 1.43

z2KfTMOV+5QYqsUPqCwEK4D7/9jGmpYywT6imGMVcuGV2J7NjOnZ/lk+l5rh9+Jtf7gX/VZmHwAyU9Qj8JtZIaWro6BaONcvcQkKm1Y475VNrfijtxBCR6p/QK/Oaud843usc8s7sNpF+uKA9FqPX/fF9E+mPpJtwQ1EU1+k8k/gRZLQEG6+kq5NFCreeojsFaZCs2r6DRfHhJLxAphuGGPTZxosUOPs2SUoGunaX9J6EL+G33/zrxJ7Fb0lc1bMIAGl9ljNCD5AaNIYbKFbLNLk+HjOnCTrhf88rjpipPwl06OK1vQUlFlqDzbrFpIe9nc4wiz9EwTbcJLpFm+FavLILYwP1WvJ4k6bUcn4YBIxXSfrT8Kygw57e9A=
Correct. Since we want to compute a probability on an individual x, we can only do this if the population distribution is Normal.
Incorrect. Since we want to compute a probability on an individual x, we can only do this if the population distribution is Normal.
2
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### Question 1.44

LlwmnlFr5Cgv7tWKmScJHvlHsYacNF7KisTyS3QogXtU0Kk5Wi1xAnUQXv5gW/DveQErB/ABAJk6SEge7JKY5TnTuOoD1XiG29042iBd7iWf6filJABJdt+ggO/8mu9QVvVmHJ5zTPT0grI2ULKjd8f8uGYUhzLtcSQWugfR02IVZSFuftxoJFt2DG1nMgjhNKQFRJGyHBA9YFG7u9gA3TxkiHIJ7QUU95RLyK2X89jeLZ39KFrHFZR2FCkpbgYDTLuTcSh5T14YZesCTl+J0kKz/6XCx3LfNTH3Ny2WZRU1eXJ1BN3bZ+Y+RfULxpXNIMz8eHnH5y4k9chLzydOCa9KxBdTat6lRTnaNTM9W//s5JQmUjoeaac0rgra6hHohEy2kSgjuuTyeWPe
Correct. Since we want to compute a probability on $$\overline{x}$$, we need the distribution of $$\overline{x}$$’s to be either Normal or approximately Normal.
Incorrect. Since we want to compute a probability on $$\overline{x}$$, we need the distribution of $$\overline{x}$$’s to be either Normal or approximately Normal.
2
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