Question

Suppose that the weight of sweet cherries is normally distributed with mean = 6oz and Standard Deviation = 1.4oz. What proportion of sweet cherries weigh more than 4.7oz? (Round your answer to 4 decimal places)

Answer 1

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The normal distribution parameters are given as:

Mean, Mu = 6 oz

Stdev, Sigma = 1.4 oz

We will these parameters along with the standardization formula to solve the problem. The formula for standardization is : Z = (X-Mean)/(Stdev)

We have been asked to find P(X>4.7) = ?

Standardizing this:

P(X>4.7) = P(Z> (4.7-6)/1.4)

= P(Z> -.9286)

= 1- P(Z<= -0.9286) [use the Z-tables to convert the Z value into a cumulative probability , it is 0.1766]

= 1-.1766

= .8234

Answer : 0.8234