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On a test with a population mean of 75 and standard deviation equal to 16, if the scores are normally distributed, what percentage of scores fall between 70 and 80?

User Mofury
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1 Answer

3 votes

Answer:

Percentage of scores that fall between 70 and 80 = 24.34%

Explanation:

We are given a test with a population mean of 75 and standard deviation equal to 16.

Let X = Percentage of scores

Since, X ~ N(
\mu,\sigma^(2))

The z probability is given by;

Z =
(X-\mu)/(\sigma) ~ N(0,1) where,
\mu = 75 and
\sigma = 16

So, P(70 < X < 80) = P(X < 80) - P(X <= 70)

P(X < 80) = P(
(X-\mu)/(\sigma) <
(80-75)/(16) ) = P(Z < 0.31) = 0.62172

P(X <= 70) = P(
(X-\mu)/(\sigma) <
(70-75)/(16) ) = P(Z < -0.31) = 1 - P(Z <= 0.31)

= 1 - 0.62172 = 0.37828

Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%

User Unnikrishnan R
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