Question

14. A common design requirement is that an environment must fit the range of people who fall between the 5th percentile for women and the 95th percentile for men. In designing an assembly work​ table, the sitting knee height must be​ considered, which is the distance from the bottom of the feet to the top of the knee. Males have sitting knee heights that are normally distributed with a mean of 21.1 in. and a standard deviation of 1.1 in. Females have sitting knee heights that are normally distributed with a mean of 19.4 in. and a standard deviation of 1.0 in. Use this information to answer the following questions.

1. What is the minimum table clearance required to satisfy the requirement of fitting​ 95% of​ men?____​(Round to one decimal place as​ needed.)

2. Determine if the following statement is true or false. If there is clearance for 95% of males, there will certainly be clearance for all women in the bottom 5%.

A. The statement is true because some women will have sitting knee heights that are outliers.

B. The statement is false because some women will have sitting knee heights that are outliers.

C. The statement is true because the 95th percentile for men is greater than the 5th percentile for women.

D. The statement is false because the 95th percentile for men is greater than the 5th percentile for women.

3. The author is writing this exercise at a table with a clearance of 23.9 in. above the floor. What percentage of men fit this table?

4. What percentage of women fit this table?

5. Does the table appear to be made to fit almost everyone? Choose the correct answer below.

A. The table will fit almost everyone except about 22% of men with the largest sitting knee heights.

B. The table will fit only 22% of men.

C. The table will only fit 1% of women.

D. Not enough information to determine if the table appears to be made to fit almost everyone.

15.The lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 15 days.

a. Find the probability of a pregnancy lasting 308 days or longer.____​(Round to four decimal places as​ needed.)

b. If the length of pregnancy is in the lowest 4​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.____​(Round to four decimal places as​ needed.)

Answer 1

1.

Z value for 95% percentile = 1.64

Minimum table clearance required to satisfy the requirement of fitting​ 95% of​ men = 21.1 + 1.64 * 1.1 = 22.9 inch

2.

Z value for 5% percentile = -1.64

5th percentile for women. = 19.4 - 1.64 * 1 = 17.8 inch

Since, 95th percentile for men is greater than the 5th percentile for women, the correct answer is,

C. The statement is true because the 95th percentile for men is greater than the 5th percentile for women.

3.

Z value for  23.9 in = ( 23.9 - 21.1) / 1.1 = 2.55

Percentile for z = 2.55 is 0.9946

Thus, 99.46% percentage of men fit this table.

4.

Z value for  23.9 in for women = ( 23.9 - 19.4) / 1 = 4.5

Percentile for z = 4.5 is 1

Thus, 100% percentage of women fit this table.

5.

A. The table will fit almost everyone except about 0.5% of men with the largest sitting knee heights.

15.

Probability of a pregnancy lasting 308 days or longer = P(X > 308) = P[Z > (308 - 269) / 15]

= P[Z > 2.6] = 0.0047

b.

Z value for 4% percentile is -1.75

The length that separates premature babies from those who are not premature = 269 - 1.75 * 15 = 242.75 days