MATH SOLVE

10 months ago

Q:
# the height of woman ages 20-29 is normally distributed , with a mean of 64.4 inches. assuming the standard diviation = 2.3 inches. are you more likely to randomly select 1 woman with a height less than 64.9 inches or are you more likely to select a sample of 22 woman with a mean height less than 64.9 inches.(USING STANDARD NORMAL TABLE)A. what is the probability of randomly selecting 1 woman with a height less than 64.9 inches?B. what is the probability of selecting a sample of 22 woman with a mean height less then 64.9 inches?C. are you more likely to randomly select 1 woman with a height less than 64.9 inches or are you more likely to select a sample of 22 woman with a mean height of 64.9 inches?

Accepted Solution

A:

Answer:a) 0.586b) 0.846c) We are more likely to select a sample of 22 woman with a mean height of 64.9 inches.Step-by-step explanation:We are given the following information in the question:
Mean, μ = 64.4 inchesStandard Deviation, σ =2.3 inchesWe are given that the distribution of height of woman is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]a) P(selecting 1 woman with a height less than 64.9 inches)P(x < 64.9)
[tex]P( x < 64.9) = P( z < \displaystyle\frac{64.9 - 64.4}{2.3}) = P(z < 0.2173)[/tex]
Calculation the value from standard normal z table, we have, [tex]P(x < 64.9) = 0.586 = 58.6\%[/tex]b) P(selecting 22 woman with a height less than 64.9 inches)Standard error due to sampling =[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{2.3}{\sqrt{22}} = 0.49[/tex]P(x < 64.9)
[tex]P( x < 64.9) = P( z < \displaystyle\frac{64.9 - 64.4}{\frac{2.3}{\sqrt{22}}}) = P(z < 1.0196)[/tex]
Calculation the value from standard normal z table, we have, [tex]P(x < 64.9) = 0.846 = 84.6\%[/tex]c) Since,P(selecting 22 woman with a height less than 64.9 inches) > P(selecting 1 woman with a height less than 64.9 inches)We are more likely to select a sample of 22 woman with a mean height of 64.9 inches.