Respuesta :
Answer:
The cutoff numbers for the number of chocolate chips in acceptable cookies are 16.242 and 27.758
Step-by-step explanation:
Problems of normally distributed samples can be 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 = 22, \sigma = 3.5[/tex]
Middle 90%
Between the 50 - (90/2) = 5th percentile to the 50 + (90/2) = 95th percentile.
5th percentile:
X when Z has a pvalue of 0.05. So X when Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 22}{3.5}[/tex]
[tex]X - 22 = -1.645*3.5[/tex]
[tex]X = 16.242[/tex]
95th percentile:
X when Z has a pvalue of 0.95. So X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 22}{3.5}[/tex]
[tex]X - 22 = 1.645*3.5[/tex]
[tex]X = 27.758[/tex]
The cutoff numbers for the number of chocolate chips in acceptable cookies are 16.242 and 27.758
Answer:
16,242 and 27,758
Step-by-step explanation:
The first thing to keep in mind is that
middle 90% means 5% bottom and 5% top are excluded.
we have that the mean (m) is 22 and the standard deviation (sd) 3,5.
using percentile table, we will note down the z score corresponding to 5th and 95th percentile
z (5th) = -1.645
z (95th) = 1,645
lower cut off value =
m + z (5th) * sd
22 + (-1,645) * 3,5
= 16,242
upper cut value
m + z (95th) * sd
22 + (1,645) * 3.5
= 27,758
Therefore the values are 16,242 and 27,758
