Question

1.The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.63 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.2 inches and a standard deviation of 2.54 inches.

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

z =

b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.

%

c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

z =

d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

%

2.A population of values has a normal distribution with μ=32.1μ=32.1 and σ=61.9σ=61.9. You intend to draw a random sample of size n=48n=48.

What is the mean of the distribution of sample means?
μ¯x=μx¯=

What is the standard deviation of the distribution of sample means?
(Report answer accurate to 2 decimal places.)
σ¯x=σx¯=

3. A population of values has a normal distribution with μ=114.5μ=114.5 and σ=47σ=47. You intend to draw a random sample of size n=147n=147.

What is the mean of the distribution of sample means?
μ¯x=μx¯=

4. What is the standard deviation of the distribution of sample means (i.e. the standard error)?
(Report answer accurate to 2 decimal places.)
σ¯x=σx¯=

5. SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?

Answer 1

Solution :

Given that ,

1) mean = = 69.3 in. (men)

standard deviation = = 2.63 in.

mean = = 64.2 in. (women)

standard deviation = = 2.54 in.

a) Using z-score formula,

x = 6 feet 3 in. = 75 in.

z = x - /

z = 75 - 69.3 / 2.63

z = 2.17

b) P(x < 75) = P[(x - ) / < (75 - 69.3) / 2.63]

= P(z < 2.17)

Using z table,

= 0.9850

The percentage is = 98.50%

c) x = 5 feet 11 in. = 71 in.

Using z-score formula,

z = x - /

z = 71 - 64.2 / 2.54

z = 2.68

d) P(x > 71) = 1 - p( x< 71)

=1- p P[(x - ) / < (71 - 64.2) / 2.54]

=1- P(z < 2.68)

Using z table,

= 1 - 0.9963

= 0.0037

The percentage is = 0.37%

2) mean = = 32.1

standard deviation = = 61.9

n = 48

=   = 32.1

= / n = 61.9 / 48 = 8.93

3) mean = = 114.5

standard deviation = = 47

n = 147

=   = 114.5

4) = / n = 47 / 147 = 3.88

5) Population standard deviation = = 300

Margin of error = E = 25

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z_/2 = 1.96

sample size = n = [Z_/2* / E] ²

n = [ 1.96 * 300 / 25 ]²

n = 553.19

Sample size = n = 554 students.