Question

Scores on a test have a mean of 72.9 and 9 percent of the scores are above 85. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth, if necessary.

Answer #1

Solution :

mean = = 72.9

standard deviation = = ?

x = 85

Using standard normal table,

P(Z > z) = 9%

1 - P(Z < z) = 0.09

P(Z < z) = 1 - 0.09

P(Z < 1.34) = 0.91

z = 1.34

Using z-score formula,

x = z * +

= (x - ) / z = (85 - 72.9) / 1.34 = 9

standard deviation = 9

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