Question

Chocolate chip cookies have a distribution that is approximately normal with a mean of 23.6 chocolate chips per cookie and a standard deviation of 2.3 chocolate chips per cookie. Find P(5) and P(95). How might those values be helpful to the producer of the chocolate chip​ cookies?

Answer 1

Those values can be helpul to find an ideal range for the number of chocolate chips per cookie.

P(5) = 19.82 chips per cookie.

P(95) = 27.38 chips per cookie.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 23.6, \sigma = 2.3`

How might those values be helpful to the producer of the chocolate chip​ cookies?

Those values can be helpul to find an ideal range for the number of chocolate chips per cookie.

P(5)

5th percentile, which is the value of X when Z has a pvalue of 0.05. So it is X when Z = -1.645.

`Z = (X - \mu)/(\sigma)`

`-1.645 = (X - 23.6)/(2.3)`

`X - 23.6 = -1.645*2.3`

`X = 19.82`

P(5) = 19.82 chips per cookie.

P(95)

95th percentile, which is the value of X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

`Z = (X - \mu)/(\sigma)`

`1.645 = (X - 23.6)/(2.3)`

`X - 23.6 = 1.645*2.3`

`X = 27.38`

P(95) = 27.38 chips per cookie.