Question

The number of raisins in a 24 oz. box of Raisin Bran Cereal is normally distributed with a mean of 100 raisins and a standard deviation of 15 raisins.

1. What is the probability that a box of cereal will have no more than 110 raisins in it?
   1. 0
   2. 0.75
   3. 0.68
   4. 0.83

2. What is the probability that a box of cereal will have between 80 to 120 raisins?
   1. 0.5
   2. 0.32
   3. 0.67
   4. 0.82

 

3. Help the manager decide the raisin count per box for the top 3% boxes with respect to the raisin content. In other words, how many raisins must be in a box so that the box is in the top 3% with respect to raisin content? Pick the answer that comes closest to the correct answer.
   1. More than 106
   2. More than 112
   3. More than 128
   4. More than 142

 

4. Assume the manager wants 0.9 probability that a box of cereal has 60 or more raisins. Also assume that the process remains normally distributed and the standard deviation of process remains the same (i.e., 15 raisins). What should be the mean of the number of raisins in a box of cereal? Round off to nearest integer.
   1. 79
   2. 85
   3. 95
   4. 105

Answer 1

Solution:

Let X be a random variable which represents the number of raisin in the box of cereal.

Given that,   i.e.    

Now we know that,

   then  .

1) We have to find P(X ≤ 110).

Using NORM.S.DIST function of excel we get,

P(Z ≤ 0.6667) = 0.7475

probability that a box of cereal will have no more than 110 raisins in it is 0.75.

2) We have to find P(80 < X < 120).

Using NORM.S.DIST function of excel we get,

P(Z < 1.3333) = 0.9088 and P(Z < -1.3333) = 0.0912

The probability that a box of cereal will have between 80 to 120 raisins is 0.82.

3) To obtain the number of raisin in the box such that box is in top 3%, we need to calculate 97th percentile. A box is in top 3% means that 97% of the boxes is under this.

Let the 97th percentile be x_{​​​​​​1} i.e. let the number of raisin in the box be x​​​​​_{​​​​​​1} so that box is in top 3%.

Hence,  

Now using excel we get, P(Z < 1.88) = 0.97

Comparing and P(Z < 1.88) = 0.97 we get,

Hence, more than 128 raisins must be in a box so that the box is in the top 3% with respect to raisin content.

4) Given that, P(X ≥ 60) = 0.9

................................(1)

Using excel we get, P(Z < -1.282) = 0.1 and comparing this expression with above expression given in (1) we get,

The mean of the number of raisins in a box of cereal is 79.