Question

1. which of the following examples would not be able to be described by a normal curve?

(a) the height of the actresses on broadway
(b) rolling two dice, 50 times, and observing the sum of the two dice
(c) how many ounces of sugar in a 20 oz container
(d) the shoe size of all the students in the school

1. in a rural part of the state, the distance to the local hospital varies according to a normal distribution, with a mean of 10.5 miles and a standard deviation of 3 miles. the graph below shows the distribution.

image of a normal curve with values 1.5, 4.5, 7.5, 10.5, 13.5, 16.5, 19.5 on the x - axis and some percentage markings on the curve
according to the empirical rule, what percentage of houses are between 4.5 miles and 13.5 miles away from the hospital?
(a) 47.5%
(b) 84%
(c) 81.6%
(d) 50%

1. the amount of weight a freshman college student gains when they attend a four - year college varies according to a normal distribution, with a mean of 2.5 lbs and a standard deviation of 10.8 lbs. if a college student lost 4 pounds, find and interpret their z - score.

(a) their z - score of - 0.602 says that they lost - 0.602 standard deviations below the mean weight.
(b) their z - score of - 0.602 says that they were 0.602 standard deviations below the mean weight gain of 2.5 lbs.
(c) their z - score of 0.139 says that they lost 0.139 standard deviations of weight.
(d) their z - score of 0.602 says that they were 0.602 standard deviations above the mean weight loss of 2.5 lbs.

Explanation:

Question 1

Step1: Recall normal curve properties

A normal curve (normal distribution) is symmetric, bell - shaped, and describes continuous data with most values near the mean.

Step2: Analyze each option

• Option A: Height of actresses is continuous and likely normally distributed (most around a mean height, symmetric).
• Option B: Rolling two dice 50 times and observing the sum. The sum of two dice has a discrete uniform - like distribution (sum ranges from 2 - 12 with specific probabilities, not a normal curve shape).
• Option C: Ounces of sugar in a 20 oz container. This is continuous (could have small variations around 20) and likely normally distributed.
• Option D: Shoe size of students. Shoe sizes are discrete but often approximately normally distributed (most around a mean size, symmetric).

Step1: Recall the Empirical Rule

The Empirical Rule states that for a normal distribution:

• Approximately 68% of data is within $\mu\pm\sigma$
• Approximately 95% of data is within $\mu\pm2\sigma$
• Approximately 99.7% of data is within $\mu\pm3\sigma$

Given $\mu = 10.5$ miles and $\sigma=3$ miles.

Step2: Find the number of standard deviations for the bounds

• For 4.5 miles: $z=\frac{4.5 - 10.5}{3}=\frac{- 6}{3}=- 2$ (2 standard deviations below the mean)
• For 13.5 miles: $z=\frac{13.5 - 10.5}{3}=\frac{3}{3} = 1$ (1 standard deviation above the mean)

Step3: Calculate the percentage

• Percentage within $\mu - 2\sigma$ to $\mu$: $\frac{95\%}{2}=47.5\%$
• Percentage within $\mu$ to $\mu+\sigma$: $\frac{68\%}{2}=34\%$
• Total percentage between 4.5 ( $\mu - 2\sigma$) and 13.5 ( $\mu+\sigma$) is $47.5\%+34\% = 81.5\%\approx81.6\%$ (due to rounding)

Step1: Recall the 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.

Here, $x=- 4$ (since the student lost 4 pounds, so the weight gain is - 4), $\mu = 2.5$ lbs, and $\sigma = 10.8$ lbs.

Step2: Calculate the z - score

$z=\frac{-4 - 2.5}{10.8}=\frac{-6.5}{10.8}\approx - 0.602$

Step3: Interpret the z - score

A negative z - score means the data point is below the mean. A z - score of - 0.602 means the student's weight gain (which is a loss of 4 pounds, so $x=-4$) is 0.602 standard deviations below the mean weight gain.

Answer:

B. Rolling two dice, 50 times, and observing the sum of the two dice

Question 2