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

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

Step-by-step 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

The probability of juniors whose score is at or below 66

                = 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

 = 0.1587

The percentage of juniors whose score is at or below 66

P(X≤ 66) = 15 %

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