Question

1. the chest sizes of a group of men are normally distributed with a mean of 36 inches and standard deviation of 2 inches. what percent of men have chest sizes more than 40 inches? 2. a competency test has scores with a mean of 80 and a standard deviation of 10. a histogram shows that the data is normally distributed. use the empirical rule to find the percent of scores between 50 and 90? 3. a normal distribution has a mean of 28 and a standard deviation of 3. which of the following intervals contains 99.7% of the distribution? i) (22,34) j) (25,31) k) (20,35) l) (19,37) a standardized test has a mean of 500 and a standard deviation of 100. kristin got 700 on her what percent of students are below her score of 700?

Explanation:

Step1: Calculate z - score for first problem

The z - score formula is $z=\frac{x-\mu}{\sigma}$. For the first problem, $\mu = 36$, $\sigma=2$, and $x = 40$. So $z=\frac{40 - 36}{2}=\frac{4}{2}=2$.

Step2: Find the area to the right of z - score

Using the standard normal distribution table, the area to the left of $z = 2$ is $0.9772$. The area to the right (percent of men with chest sizes more than 40 inches) is $1-0.9772 = 0.0228$ or $2.28\%$.

Step3: Calculate z - scores for second problem

For $x_1 = 50$, $z_1=\frac{50 - 80}{10}=\frac{- 30}{10}=-3$. For $x_2 = 90$, $z_2=\frac{90 - 80}{10}=\frac{10}{10}=1$.

Step4: Use the Empirical Rule

The Empirical Rule states that about $68\%$ of the data is within 1 standard - deviation of the mean, about $95\%$ is within 2 standard - deviations, and about $99.7\%$ is within 3 standard - deviations. The area between $z=-3$ and $z = 1$: The area to the left of $z=-3$ is $0.0015$, and the area to the left of $z = 1$ is $0.8413$. So the area between them is $0.8413-0.0015 = 0.8398$ or $83.98\%$.

Step5: Recall the Empirical Rule for third problem

The Empirical Rule states that $99.7\%$ of the data in a normal distribution is within $z=-3$ and $z = 3$ of the mean. If $\mu = 28$ and $\sigma = 3$, then the interval is $(\mu-3\sigma,\mu + 3\sigma)=(28-3\times3,28 + 3\times3)=(19,37)$.

Step6: Calculate z - score for fourth problem

For $x = 700$, $\mu = 500$, $\sigma = 100$, $z=\frac{700 - 500}{100}=\frac{200}{100}=2$. The area to the left of $z = 2$ (percent of students below 700) is $0.9772$ or $97.72\%$.

Answer:

1. $2.28$
2. $83.98$
3. l) $(19,37)$
4. $97.72$