Question

A particular fruit's weights are normally distributed, with a mean of 694 grams and a standard deviation of 40 grams. The heaviest 2% of fruits weigh more than how many grams? Give your answer to the nearest gram.

Answer 1

Let X = fruit's weight Here X ~ N(694, 40). P(X > c) = 0.02 = P(X<c) = 0.98 = c = 776 (U se R code: round(qnorm(0.98, 694, 40))). Answer: 776 gm.

Answer 2

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Answer 3

SOLUTION :

Let a grams be the weight  over which 2% fruits are there.

P( x > a) = P( z > (a - m)/s)

=> 0.02 =  P( z > (a - m)/s)

From ND table : 

z > 2.054

The cutoff point is z = 2.054

So, 

a = z * s + m = 2.054 * 40 + 694  = 776.16 = 776 grams

So, the heaviest 2% fruits weigh more than 776 grams (ANSWER).