Question

math data financial
chapter 1 - 4 problem solving
use the z - score formula to complete the following questions. use a separate sheet to compute the data and place the answers on the answer sheet.

1. a summer camp is taking their 220 sixth graders on a trip to an amusement park. for safety purposes, some of the rides have height requirements. the campers’ heights have a mean of 56 inches and a standard deviation of 3 inches. what is the z - score for a camper with a height of 62 inches?
2. a camper on the trip has a height of 54 inches. express his height as a z - score. round to the nearest hundredth?
3. the height of a certain student on this trip had a z - score of - 0.5. what is the student’s height in inches?
4. the distribution of money the campers spent on souvenirs has mean $12 and standard deviation $1.20. find the amount spent by a specific camper who had a z - score of 1.5?
5. recall the amusement park trip from questions 1 - 4. a certain ride requires riders to be at least 51 inches tall (raw score). the heights are normally distributed with mean 56 and standard deviation 3. approximately how many of the camp’s 220 sixth graders will not be allowed on the ride? first, on the answer sheet, draw a sketch of the bell curve. label where the mean is and where the height of 51 could be. shade the area under the curve below 51 inches which will represent the percent of campers who are shorter than 51 inches.
6. convert the 51 inches in #5 to a z - score, then use the table to find the percentage of students that will not be allowed on the ride.
7. use the data from questions 5 and 6 to calculate if another ride requires riders to be 60 inches tall, what percentage of the campers will be able to go on the ride.
8. families of students at smithtown high school were surveyed about their vacation expenses. the results were normally distributed with mean $2,313 and standard deviation $390. what percent of the families took vacations that cost between $2,000 and $3,000?
9. use the data from the families’ vacation expenses in question #8 to find the percentage of families that had vacation expenses between $1,000 and $2,000. round to the nearest percent.

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the data point, $\mu$ is the mean and $\sigma$ is the standard deviation.

Step2: Solve question 1

Given $\mu = 56$, $\sigma=3$ and $x = 62$. Substitute into the formula: $z=\frac{62 - 56}{3}=\frac{6}{3}=2$.

Step3: Solve question 2

Given $\mu = 56$, $\sigma = 3$ and $x = 54$. Substitute into the formula: $z=\frac{54 - 56}{3}=\frac{-2}{3}\approx - 0.67$.

Step4: Solve question 3

Given $z=-0.5$, $\mu = 56$, $\sigma = 3$. Rearrange the z - score formula $z=\frac{x-\mu}{\sigma}$ to $x=z\sigma+\mu$. Then $x=-0.5\times3 + 56=-1.5+56 = 54.5$.

Step5: Solve question 4

Given $z = 1.5$, $\mu=12$, $\sigma = 1.2$. Rearrange the z - score formula to $x=z\sigma+\mu$. Then $x=1.5\times1.2+12=1.8 + 12=13.8$.

Step6: Solve question 5 and 6

For $x = 51$, $\mu = 56$, $\sigma = 3$. Calculate the z - score: $z=\frac{51 - 56}{3}=\frac{-5}{3}\approx - 1.67$. Looking up in the standard normal table, the area to the left of $z=-1.67$ is approximately $0.0475$. The number of campers not allowed on the ride is $0.0475\times220\approx10.45\approx10$.

Step7: Solve question 7

For $x = 60$, $\mu = 56$, $\sigma = 3$. Calculate the z - score: $z=\frac{60 - 56}{3}=\frac{4}{3}\approx1.33$. Looking up in the standard normal table, the area to the left of $z = 1.33$ is approximately $0.9082$. So the percentage of campers able to go on the ride is $0.9082\times100\% = 90.82\%\approx91\%$.

Step8: Solve question 8

For $x_1 = 2000$, $\mu = 2313$, $\sigma = 390$, $z_1=\frac{2000 - 2313}{390}=\frac{-313}{390}\approx - 0.80$. For $x_2 = 3000$, $\mu = 2313$, $\sigma = 390$, $z_2=\frac{3000 - 2313}{390}=\frac{687}{390}\approx1.76$. The area between $z=-0.80$ and $z = 1.76$ is $0.9608-0.2119 = 0.7489\approx75\%$.

Step9: Solve question 9

For $x_1 = 1000$, $\mu = 2313$, $\sigma = 390$, $z_1=\frac{1000 - 2313}{390}=\frac{-1313}{390}\approx - 3.37$. For $x_2 = 2000$, $\mu = 2313$, $\sigma = 390$, $z_2=\frac{2000 - 2313}{390}\approx - 0.80$. The area between $z=-3.37$ and $z=-0.80$ is $0.2119 - 0.0004=0.2115\approx21\%$.

Answer:

1. $z = 2$
2. $z\approx - 0.67$
3. $x = 54.5$ inches
4. $\$13.8$
5. Approximately 10 campers
6. $z\approx - 1.67$, approximately $4.75\%$
7. Approximately $91\%$
8. Approximately $75\%$
9. Approximately $21\%$