Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 23.6, \sigma = 2.3[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 23.6}{2.3}[/tex]
[tex]X - 23.6 = -1.645*2.3[/tex]
[tex]X = 19.82[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 23.6}{2.3}[/tex]
[tex]X - 23.6 = 1.645*2.3[/tex]
[tex]X = 27.38[/tex]
P(95) = 27.38 chips per cookie.
