• Step 1 Determine $X$, $\mu$, and $\sigma$.

  We are given that the normal random variable $X$ = Math SAT score has mean $\mu =514$ and standard deviation $\sigma =118$. In the center of the number line, mark the mean $\mu =514$. Also mark on the number line the value of $X$ about which the problem is asking. Figure 48 shows the graph of $X$ (the Math SAT scores) with the mean of 514 and the score of 396 marked.

  You need to know the proportion of scores below 396, so shade the area under the curve to the left of 396. We can express this proportion as a probability, the probability that a randomly chosen student will score less than 396, or $P(X<396)$. Just by looking at Figure 48, you should be able to get a rough idea of what the proportion of these scores will be. Certainly, this proportion will be less than 50%. If you get an answer such as “60%” for your proportion, you should recognize that it is wrong.

  

  FIGURE 48 Graph of proportion of Math SAT scores lower than 396.

• Step 2 Standardize.

  Now standardize the random variable $X$ to the standard normal $Z$:

  $Z=\frac{X-\mu }{\sigma }=\frac{X-514}{118}$

  Find the $Z$-value corresponding to the Math SAT score of 396:

  $\frac{396-\mu }{\sigma }=\frac{396-514}{118}=-1$

  So the $Z$-value associated with a score of 396 is −1, which indicates that the score of 396 is 1 standard deviation below the mean of 514.

  

  FIGURE 49 Graph of $P(Z<-1)$.

• Step 3 Draw the standard normal curve.

  Scores less than 396 are more than 1 standard deviation below the mean, so shade the area to the left of −1 in Figure 49. Now find the area to the left of $Z=-1$ using the methods of Section 6.4.

  Step 4 Find the area under the standard normal curve.

  The $Z$ table tells us that the area to the left of $Z=-1$ is 0.1587.

  The proportion of scores below 396 is 0.1587, or 15.87%. Note that this value for $P(X<396)$ agrees with our earlier intuition that the proportion was less than 50%.

NOW YOU CAN DO

• Step 1 Determine $X$, $\mu$, and $\sigma$.

  We have already seen that $X=\text{Math SAT score}$ has mean $\mu =514$ and standard deviation $\sigma =118$. Once again, draw a graph of the distribution of scores $X$, with the mean 514 in the middle, the score 215 to the left of the mean, and the score 595 to the right of the mean, as in Figure 50.

• Step 2 Standardize.

  This is a “between” example, where two values of $X$ are given, and we are asked to find the area between them. In this case, just standardize both of these values of $X$ to get a $Z$-value for each:

  $\begin{array}{ccc}Z=\frac{215-\mu }{\sigma }=\frac{215-514}{118}\approx -2.53& \text{and}& Z=\frac{595-\mu }{\sigma }=\frac{595-514}{118}\approx 0.69\end{array}$

  FIGURE 50 Graph of percentage of students scoring between 215 and 595 on the Math SAT.

  

• Step 3 Draw the standard normal curve.

  Draw a graph of $Z$, shading the area between $Z=-2.53$ and $Z=0.69$, as shown in Figure 51. Again, the key is that the area between $Z=-2.53$ and $Z=0.69$ is exactly the same as the area between $X=215$ and $X=595$.

• Step 4 Find area under the standard normal curve.

  Figure 51 is a Case 3 problem from Table 8 (page 355). Find the area to the left of 0.69, which is 0.7549, and the area to the left of −2.53, which is 0.0057. Subtracting the smaller from the larger gives us

  $P(-2.53<Z<0.69)=0.7549-0.0057=0.7492$

  Thus, the percentage of Math SAT scores that are between 215 and 595 is 74.92%.

  

  FIGURE 51 Graph of percentage of $Z$ - values between −2.53 and 0.69.

NOW YOU CAN DO

• Step 1 Determine $X$, $\mu$, and $\sigma$, and draw the normal curve for $X$.

  We already know that $X=\text{Math SAT score}$, $\mu =514$, and $\sigma =118$. The value of $X$ in which we are interested refers to high scores, so that $X_1$ will be at the far right of the distribution of $X$. Only 1% of scores will be greater than this score, so the area to the right of $X_1$ is 0.01, as shown in Figure 52.

  FIGURE 52 $X_1$ is the cutoff value (or critical value) of $X$, at which students will win a fellowship to an Ivy League university.

  
  Page 373

• Step 2 Find the $Z$-value corresponding to the desired area.

  The area to the right of $X_1$ equals 0.01, so that the area to the left of $X_1$ equals $1-0.01=0.99$. Looking up 0.99 on the inside of the $Z$ table gives us $Z=2.33$.

• Step 3 Transform using the formula $X_1=Z\sigma +\mu$.

  We calculate

  $X_1=Z\sigma +\mu =\left(2.33\right)\left(118\right)+514=788.94$

NOW YOU CAN DO

NOW YOU CAN DO

1. Figure 55 shows that $P(27,000\le X\le 30,000)=0.8570111819\approx 0.857$.
2. Figure 56 shows that the value for $X_1$, such that $P(X\le X_1)=0.99$, is given by $X_1=\$31,046.34788\approx \$31,046.35$.
   

   FIGURE 55 TI-83/84: Finding a probability.
   

   FIGURE 56 TI-83/84: Finding a value of $X$.

1. Excel provides the cumulative probabilities $P(X\le 27,000)$ in Figure 57 and $P(X\le 30,000)$ in Figure 58. To find $P(X\le 27,000\le 30,000)$, we subtract $P(X\le 27,000)$ from $P(X\le 30,000)$: $P(27,000\le X\le 30,000)=0.899727432-0.042716221=0.857011211\approx 0.8570$
   

   FIGURE 57 Excel: $P(X\le 27,000)$.
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   FIGURE 58 Excel: $P(X\le 30,000)$.
2. Excel provides the result shown in Figure 59: $X_1=\$31,046.34788\approx \$31,046.35$
   

   FIGURE 59 Excel: Finding a value of $X$.

1. Similar to Excel, Minitab asks you to take the difference of two cumulative probabilities: $P(X\le 27,000)$ in Figure 60 and $P(X\le 30,000)$ in Figure 61: $P(27,000\le X\le 30,000)=0.899727-0.0427162=0.8570108\approx 0.8570$
   

   FIGURE 60 Minitab: $P(X\le 27,000)$.
   

   FIGURE 61 Minitab: $P(X\le 30,000)$.
2. The results are given in Figure 62: $X_1=\$31,046.30$
   

   FIGURE 62 Minitab: Finding a value of $X$.

1. JMP also asks you to take the difference of two cumulative probabilities: $P(X\le 27,000)$ in Figure 63 and $P(X\le 30,000)$ in Figure 64: $P(27,000\le X\le 30,000)=0.899727432-0.0427162208=0.8570112112\approx 0.8570$
   

   FIGURE 63 JMP: $P(X\le 27,000)$.
   

   FIGURE 64 JMP: $P(X\le 30,000)$.
2. The results are given in Figure 65: $X_1=\$31,046.30$
   

   FIGURE 65 JMP: Finding a value of $X$.

NOW YOU CAN DO