# Answer to Question #2825 in Statistics and Probability for alyssa

Question #2825

Suppose the weight of full-grown oranges in a citrus grove are normally distributed with a mean of 7.3 ounces and standard deviation of 2.1 ounces. Determine the probablity that 50 oranges randomly picked from the grove have a mean weight greater than 10.5 ounces.

Expert's answer

The sample mean weight X has normal distribution

N(7.3, 2.1/√50)=& N(7.3, 0.297)

Hence

Z=(X-7.3)/0.297

has normal distribution N(0,1).

Therefore

P(X > 10.5) = P(Z > (10.5-7.3)/0.297& ) =& P(Z>10.77)& = 1 - P(Z<10.77) =& 1- F(10.77) =0.99999

N(7.3, 2.1/√50)=& N(7.3, 0.297)

Hence

Z=(X-7.3)/0.297

has normal distribution N(0,1).

Therefore

P(X > 10.5) = P(Z > (10.5-7.3)/0.297& ) =& P(Z>10.77)& = 1 - P(Z<10.77) =& 1- F(10.77) =0.99999

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