Question

1. professor atch administers two versions of the math 101 final exam, both covering the same content but within different context and numbers. version a had a standard deviation of 7 and a mean score of 78. version b had a mean score of 83 with a standard deviation of 5.8.

elijah got an 83 on version a and tanya took version b earning an 87. which of the following statements is true?
(a) tanya did better on the final exam because her score of 87 is higher than elijah’s score of 83.
(b) elijah did better on the final exam because his standardized score of 0.71 is higher than tanya’s standardized score of 0.69.
(c) tanya did better on the final exam because her standardized score of 1.29 is higher than elijah’s standardized score of 0.
(d) elijah did better on the final exam because his standardized score in the 71st percentile is higher than tanya’s standardized score in the 69th percentile.

1. the heights of nfl football players are approximately normally distributed with a mean of 71.5 inches and a standard deviation of 2.3 inches. aaron rogers is 6 feet 2 inches. what percentile does that place him at?

(a) 43rd (b) 50th (c) 83rd (d) 86th

1. the area under the standard normal curve corresponding to -0.3 < z < 1.6 is

(a) 0.327 (b) 0.471 (c) 0.563 (d) 0.954

1. the wait time at your local mcdonalds varies according to a normal distribution, with a mean of 7 minutes and standard deviation of 1.35 minutes. approximately what proportion of their wait times are over 9 minutes?

(a) 0.033
(b) 0.069
(c) 0.189
(d) 0.191

1. your favorite bag of candy says that it should have 8oz of the candy in the bag. you weigh the contents and find that you have 7.8oz. you log onto the candys website to write a strongly worded email, and see that they have a disclaimer that the weights of the bags will vary according to a normal distribution with a weight of 8oz and a standard deviation of 0.25oz.. which of the following is an appropriate conclusion based on this information?

(a) 7.8oz is still a reasonable weight of your candy bag, since it falls 0.8oz below the mean.
(b) 7.8oz is still a reasonable weight of your candy bag, since it falls 0.8 standard deviations below the me
(c) 7.8 is not a reasonable weight of your candy bag, since it falls below the mean.
(d) 7.8 is not a reasonable weight of your candy bag, since it falls above the mean.

Explanation:

Question 4

Step1: Calculate Elijah's z - score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the raw score, $\mu$ is the mean, and $\sigma$ is the standard deviation. For Elijah (Version A): $x = 83$, $\mu=78$, $\sigma = 7$. So $z_{Elijah}=\frac{83 - 78}{7}=\frac{5}{7}\approx0.71$.

Step2: Calculate Tanya's z - score

For Tanya (Version B): $x = 87$, $\mu = 83$, $\sigma=5.8$. So $z_{Tanya}=\frac{87 - 83}{5.8}=\frac{4}{5.8}\approx0.69$.

Step3: Compare z - scores

Elijah's z - score ($0.71$) is higher than Tanya's z - score ($0.69$). So Elijah did better.

Step1: Convert height to inches

Aaron Rogers' height is 6 feet 2 inches. Since 1 foot = 12 inches, 6 feet 2 inches $=6\times12 + 2=74$ inches.

Step2: Calculate z - score

Using the z - score formula $z=\frac{x-\mu}{\sigma}$, where $x = 74$, $\mu=71.5$, $\sigma = 2.3$. So $z=\frac{74 - 71.5}{2.3}=\frac{2.5}{2.3}\approx1.09$.

Step3: Find percentile from z - score

Looking up the z - score of $1.09$ in the standard normal table, the area to the left (percentile) is approximately 0.8621, which is close to the 86th percentile.

Step1: Find area to the left of $z = 1.6$

Using the standard normal table, the area to the left of $z = 1.6$ is $A_1=0.9452$.

Step2: Find area to the left of $z=- 0.3$

Using the standard normal table, the area to the left of $z=-0.3$ is $A_2 = 0.3821$.

Step3: Calculate the area between $- 0.3$ and $1.6$

The area between $-0.3$ and $1.6$ is $A_1 - A_2=0.9452-0.3821 = 0.5631\approx0.563$.

Answer:

B. Elijah did better on the final exam because his standardized score of 0.71 is higher than Tanya’s standardized score of 0.69.

Question 5