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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

Respuesta :

Answer:

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

Step-by-step 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:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

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

P(X > x)  

[tex]P( X > x) = P( z > \displaystyle\frac{x - 100}{10})=0.05[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{x - 100}{10})=0.05[/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 100}{10})=0.95 [/tex]  

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

[tex]P(z<1.645) = 0.95[/tex]

[tex]\displaystyle\frac{x - 100}{10} = 1.645\\x =116.45[/tex]  

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

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