Respuesta :
Answer:
a) 0.018
b) 0
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 14.4 in
Standard Deviation, σ = 1 in
We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(breadth will be greater than 16.5 in)
P(x > 16.5)
[tex]P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)[/tex]
[tex]= 1 - P(z \leq 2.1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%[/tex]
0.018 is the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.
b) P( with 123 randomly selected men, these men have a mean hip breadth greater than 16.5 in)
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
P(x > 16.5)
[tex]P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)[/tex]
[tex]= 1 - P(z \leq 23.29)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 16.5) = 1 - 1 = 0[/tex]
There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in
The probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in. is 0.0179.
What is a Z-table?
A z-table also known as the standard normal distribution table, helps us to know the percentage of values that are below (or to the left of the Distribution) a z-score in the standard normal distribution.
As the distribution is normally distributed, with a mean of 14.4 In., while the standard deviation is 1 inch. Therefore,
A.) The probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.
[tex]\begin{aligned}P(X > 16.5)&=P(z > 16.5)\\\\&=1-P(z < 16.5)\\\\&=1-P(\dfrac{16.5-\mu}{\sigma})\\\\&=1-P(\dfrac{16.5-14.4}{1})\\\\&=1-0.9821\\\\&=0.0179\end{aligned}[/tex]
Hence, the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in. is 0.0179.
B.) The probability that 123 men have a mean hip breadth greater than 16.5 in.
[tex]\begin{aligned}P(X > 16.5) &=P(z > \dfrac{16.5-14.4}{\frac{1}{\sqrt{123}}})\\&=P(z > 23.29)\\\\&=1-P(z\leq 23.29)\\\\&=1-1\\\\&=0\end{aligned}[/tex]
Hence, the probability that 123 men have a mean hip breadth greater than 16.5 in. is 0.
Learn more about Z-table:
https://brainly.com/question/6096474
