Answer the following.

Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 67 inches and standard deviation 6 inches.

(a)

What is the probability that an 18-year-old man selected at random is between 66 and 68 inches tall? (Round your answer to four decimal places.)

 

(b)

If a random sample of sixteen 18-year-old men is selected, what is the probability that the mean height x is between 66 and 68 inches? (Round your answer to four decimal places.)

Answer 1

Hi Tamera,

(a) This is similar to the other problems you posted. Z = (x - mu)/sigma. I will again use z₁ and z_2.

mu=67

sigma=6

x₁=66

x₂=68

Z₁ = (66 - 67) / 6 = -0.17

Z₂ = (68-67) / 6 = 0.17

From z-table:

P(Z < -0.17) = 0.4325

P(Z< 0.17) = 0.5675

Now, you want the area between the two z-scores, so we subtract:

P = 0.5675 - 0.4325 = 0.1350

(b) This is slightly different since we are now working with a sample. We need a new metric here called standard error. This is:

SE = sigma/ sqrt(n)

sigma=standard deviation

n=sample size

Here<

sigma=6

n=16

SE= 6/4 = 3/2

Now, the formula for the z-score in a sample:

z= (xbar - mu)/SE

xbar=sample mean

mu=population mean

SE=standard error

Here, I will again use z1 and z2:

xbar₁= 66

xbar₂= 68

mu=67

SE= 3/2 = 1.5

z₁ = (66-67) / 1.5 = -0.67

z₂ = (68 -67) / 1.5 = 0.67

Now, from the z-table:

P(Z< -0.67) = 0.2514

P(Z< 0.67) = 0.7486

We want the area between, so we subtract:

P = 0.7486 - 0.2514 = 0.4972

I hope this helps.

Answer 2

a)

66 to 68 inches are within 1 inch of the population mean of 67

if the population standard deviation= 6 inches

then z= 1/6 = .1667

probability = .1324 that a randomly selected 18 year old is between 66 and 68 inches

13.24% chance he's between 66 and 68 inches tall

b) with a sample of 16,

sample standard deviation = popsd/sqr(n) = 6/sqr16 = 6/4 = 1.5

66 to 67 = 1 inch

z= 1/1.5 = 2/3 = .6667

probability = .4950 that -z<x<z

49.50% chance the sample mean is between 66 and 68 inches with sample size = 16