Question

3.14 Find the 5D: Find the standard deviation of the distribution in the following situations. Alound your answers to f decimal place. a) MEN5A is an orsanization whose members have 10 in the top 2% of the population. IQs are normally distributed with mean 102 , and the minimum IQ scere fequired for admission to MENSA is 133. What is the standard deviation? b) Chotesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 187 milligrams per decititer (mg/dil). Women with chotesterol levels above 222mg/dl are considered to have high cholesteroland about 18,5% of women fall into this category. What is the standard deviation? 3.7 LA weather, Part 1: The average daily high temperature in June in LA is 78∘

F with a standard deviation of 5∘
F. Suppose that the temperatures in June closely follow a normal distribution. a) What is the probablity of observing an 83∘
F temperature. or higher in LA during a randomiy chosen day in Jine? Round your answer to 4 decimal places. b) How cool are the coldest 108 of the davs (davs with lowest average high temperature) during June in LA? Round your answer to 1 decimal place.

Answer 1

For question a, the standard deviation is approximately 15.12. For question b, the standard deviation is approximately 35.35.

Step-by-step explanation:

For question a:

a) MEN5A is an organization whose members have IQs in the top 2% of the population. IQs are normally distributed with a mean of 102 and the minimum IQ score required for admission to MENSA is 133. To find the standard deviation, we can use the formula:

SD = (X - μ) / z

Where SD is the standard deviation, X is the IQ score, μ is the mean, and z is the z-score.

Using the given data, to find the standard deviation:

SD = (133 - 102) / z

Since MENSA members have IQs in the top 2% of the population, we can use the z-score table to find the z-value corresponding to the top 2%.

From the z-score table, the z-value for the top 2% is approximately 2.05.

Substituting the values:

SD = (133 - 102) / 2.05

Simplifying:

SD = 31 / 2.05

SD ≈ 15.12 (rounded to two decimal places)

Therefore, the standard deviation for the given situation is approximately 15.12.

For question b:

b) The cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with a mean of 187 mg/dl and a standard deviation that we need to find. About 18.5% of women have high cholesterol levels above 222 mg/dl.

We can use the z-score formula to find the standard deviation:

z = (X - μ) / σ

To find the standard deviation, we need to find the X value corresponding to the 18.5% percentile using the z-score table.

From the z-score table, the z-value for the 18.5% percentile is approximately -0.99.

Substituting the values:

-0.99 = (222 - 187) / σ

Simplifying:

-0.99σ = 35

σ ≈ -35 / -0.99

σ ≈ 35.35 (rounded to two decimal places)

Therefore, the standard deviation for the given situation is approximately 35.35.