Question

Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 women are randomly selected, find the probability that they have a mean height greater than 63.0 inches.

Answer 1

Step-by-step explanation:

Given that mean,

μ

= 63.6, standard deviation,

σ

= 2.5 and sample size,

n

= 100

The respective Z-score with sample mean,

¯

X

= 63 is

Z

=

¯

X

−

μ

σ

/

√

n

Z

=

63

−

63.6

2.5

/

√

100

Z

= -2.4

Now, we use normal tables to find the probability greater than -2.4

P (Z > -2.4) = 1 - 0.0082

P (Z > -2.4) = 0.9918

Answer 2

Answer:

P (Z > -2.4) = 1 - 0.0082

P (Z > -2.4) = 0.9918

Step-by-step explanation:

Given that mean,

μ = '63.6' , standard deviation,

σ= '2.5' and sample size,

n= '100'

The respective Z-score with sample mean,

¯X = '63' is

Z = '¯X−μσ/√n'

Z = 63−63.6 ×2.5/√100

Z = -2.4

Now, we use normal tables to find the probability greater than -2.4

P (Z > -2.4) = 1 - 0.0082

P (Z > -2.4) = 0.9918