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Suppose that the scores on a statewide standardized test are normally distributed with a mean of 64 and a standard deviation of 2. Estimate the percentage of scores that were (a) between 60 and 68.

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Answer:

95. 45%

Step-by-step explanation:

Given :

Mean score, = 64

Standard deviation = 2

Score between 60 and 68

P(Z < (x - mean) / standard deviation)

P(Z < (60 - 64) / 2) = P(Z < - 2) = 0.02275 (Z probability calculator)

P(Z < (68 - 64) / 2) = P(Z < 2) = 0.97725 (Z probability calculator)

P( score between 60 and 68)

P(Z < 2) - P(Z < - 2)

0.97725 - 0.02275 = 0.9545

Score between 60 and 68 = 0.9545 = 0.9545 * 100% = 95.45%

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