Question

A physical fitness researcher devises a test of strength and finds that the scores are Normally distributed with a mean of 100 lbs and a standard deviation of 10 lbs. What is the minimum score needed to be stronger than all but 5% of the population

Answer 1

116.45 is the minimum score needed to be stronger than all but 5% of the population.

Explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 10

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

We have to find the value of x such that the probability is 0.05

P(X > x)

`P( X > x) = P( z > \displaystyle(x - 100)/(10))=0.05`

`= 1 -P( z \leq \displaystyle(x - 100)/(10))=0.05`

`=P( z \leq \displaystyle(x - 100)/(10))=0.95`

Calculation the value from standard normal z table, we have,

`P(z<1.645) = 0.95`

`\displaystyle(x - 100)/(10) = 1.645\\x =116.45`

Hence, 116.45 is the minimum score needed to be stronger than all but 5% of the population.