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

Answer:

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.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$-1.645 = \frac{X - 23.6}{2.3}$

$X - 23.6 = -1.645*2.3$

$X = 19.82$

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.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 23.6}{2.3}$

$X - 23.6 = 1.645*2.3$

$X = 27.38$

P(95) = 27.38 chips per cookie.