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The scores on a test given to all juniors in a school district are normally distributed with a mean of 73 and a standard deviation of 7. Find the percent of juniors whose score is below 66. The percent of juniors whose score is at or below 66 is

User Zhianc
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Answer:

The probability of juniors whose score is at or below =0.1587

The percentage of juniors whose score is at or below 66

P(X≤ 66) = 0.1587 = 15 percentage

Explanation:

Explanation:-

Given mean of the Population (μ) = 73

Given standard deviation of the Population(σ) = 7

Let 'X' be the random variable in normal distribution


Z = (x^(-)-mean )/(S.D) = (66-73)/(7) = -1

The probability of juniors whose score is at or below 66


P(X\leq 66 ) = P(Z\leq -1)

= 1 - P( Z>Z₁ )

= 1-( 0.5 +A( Z₁)

= 0.5 - A(-1)

= 0.5 - A(1) (∵ A(-1) = A(1))

= 0.5 - 0.3413

= 0.1587

The probability of juniors whose score is at or below 66


P(X\leq 66 ) = P(Z\leq -1) = 0.1587

The percentage of juniors whose score is at or below 66

P(X≤ 66) = 15 %

User Weaknespase
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