Question

Assume that the weights of bananas follow a Normal distribution with a mean of 121 grams, and a standard deviation of 17 grams Suppose that a store receives weekly shipments of 6 Bananas. The store owner will reject the shipment if the average weight is below 117 53g What is the probability that the next shipment of Bananas is accepted by the store owner? Select one a 0.3085 ob We cannot answer this question with the information given c0 1103 d. 0 5000 e 0.6915

Answer 1

Weight of bananas follow Normal distribution with mean 121 gram and standard deviation of 17 grams .

Thus X~ N ( $\mu$ =121 , $\sigma$ =17 )

The store receives weekly shipment of 6 bananas . Thus n = 6

The store owner reject the shipment if the average weight is below 117.53 g

Thus Probability of Rejecting shipment is given by P ( $\bar{X}$ < 117.53 )

Now we need to find the Probability that the next shipment of bananas is accepted .

So required probability is given by

Pr( shipment of bananas is accepted ) = 1 - Pr ( shipment of bananas is Rejected )

                                                        = 1 - P ( $\bar{X}$ < 117.53 )

                                                        = 1 - P ( $\frac{\bar{X}-\mu}{\sigma /\sqrt{n}}$ < $\frac{117.53-\mu}{\sigma /\sqrt{n}}$ )

                                                        = 1 - P ( Z < $\frac{117.53-121}{17 /\sqrt{6}}$ )

Pr( shipment of bananas is accepted ) = 1 - P ( Z < -0.4999841 )

where Z ~ N(0,1)

Now P ( Z < -0.4999841 ) can be obtained from standard normal probability table or more accuratly from any software like R/Excel

From R

> pnorm(Z)              # P ( Z < -0.4999841 )
[1] 0.3085431

Hence P ( Z < -0.4999841 ) = 0.3085431

Pr( shipment of bananas is accepted ) = 1 - P ( Z < -0.4999841 )

                                                        = 1 - 0.3085431 = 0.6914569

Pr( shipment of bananas is accepted ) = 0.6914569 $\approx$ 0.6915

So correct option is

Option e) 0.6915