Question

1. Lloyd’s Cereal Company packages cereal in 1-pound boxes (1 pound = 16 ounces). It is assumed that the amount of cereal per box varies according to a normal distribution with a mean of 1 pound and a standard deviation of 0.08 pound. One box is selected at random from the production line every hour, and if the weight is less than 14 ounces, the machine is adjusted to increase the amount of cereal dispensed. The probability that the amount dispensed per box will have to be increased during a 1-hour period is

A) 0.3944

B) 0.0594.

C) 0.1056.

D) 0.6056.

2. Lloyd’s Cereal Company packages cereal in 1-pound boxes (1 pound = 16 ounces). It is assumed that the amount of cereal per box varies according to a normal distribution. A sample of 25 boxes is selected at random from the production line every hour, and if the average weight is less than 14 ounces, the machine is adjusted to increase the amount of cereal dispensed. If the mean for an hour is 1 pound and the standard deviation is 0.4 pound, the probability that the amount dispensed per box will have to be increased is

A) 0.0594.

B) 0.0062.

C) 0.9401.

D) 0.9938.

Answer 1

1. Here it is assumed that the amount of cereal per box varies according to a normal distribution with a mean of 1 pound =16 ounces and a standard deviation of 0.08 pound

We need to find

Also as mean is 1pound=16 ounces and standard deviation is 0.08 pound = 1.28 ounces

As distribution is normal we can convert x to z

So answer is B. 0.0594

2. Here mean=16 ounces and standard deviation is 0.4*16=6.4

Now for sample of size n=25 and we need to find

As population is normal so as per central limit theorem sample mean is also normal and so we can convert sample mean to z

So answer is a. 0.0594