Question

The mean weight of a pineapple in a very large shipment is 5 pounds. Ten percent of these pineapples weigh less than 4 pounds. Assuming that the weights are normally distributed, what is the standard deviation in the shipment?

Answer 1

Let X be the random variable that denotes the weight of the pineapples.

Given, X is normally distributed.

The mean weight = 5 and let the standard deviation be denoted by .

Given, 10% of the pineapples weigh less than 4 pounds.

P(X < 4) = 0.10

P(Z < (4 - 5) / ) = 0.10

P(Z < -1/) = 0.10

1 - P(Z > -1/) = 0.10

P(Z > -1/) = 0.90

P(Z < 1/) = 0.90     ...........( P(Z > -a) = P(Z < a))

From standard normal table, P(Z < 1.28) = 0.90

1 / = 1.28

= 0.78

Therefore, the standard deviation is 0.78