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Question 9 A particular fruits weights are normally distributed, with a mean of 792 grams and a standard deviation of 36 gra
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Answer #1

Solution:

Given that,

mean = \mu =792

standard deviation = \sigma = 36

Using standard normal table,

P( Z > z) = 12%

P(Z > z) = 0.12

1 - P( Z < z) = 0.12

P(Z < z) = 1 - 0.12

P(Z < z) = 0.88

z = 1.17

Using z-score formula,

x = z * \sigma + \mu

x = 1.17 * 36 + 792

x = 834.12

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