Respuesta :
Answer:
a) 85.36% probability that a randomly selected bag contains between 1100 and 1500 chocolate chips
b) 16.35% probability that a randomly selected bag contains fewer than 1125 chocolate chips
c) 0.3446 = 34.46% of bags contains more than 1200 chocolate chips
d) 91st percentile.
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 question, we have that:
[tex]\mu = 1252, \sigma = 129[/tex]
(a) What is the probability that a randomly selected bag contains between 1100 and 1500 chocolate chips?
This is the pvalue of Z when X = 1500 subtracted by the pvalue of Z when X = 1100.
X = 1500
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1500 - 1252}{129}[/tex]
[tex]Z = 1.92[/tex]
[tex]Z = 1.92[/tex] has a pvalue of 0.9726
X = 1100
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1100 - 1252}{129}[/tex]
[tex]Z = -1.18[/tex]
[tex]Z = -1.18[/tex] has a pvalue of 0.1190
0.9726 - 0.1190 = 0.8536
85.36% probability that a randomly selected bag contains between 1100 and 1500 chocolate chips
(b) What is the probability that a randomly selected bag contains fewer than 1125 chocolate chips?
This is the pvalue of Z when X = 1125. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1125 - 1252}{129}[/tex]
[tex]Z = -0.98[/tex]
[tex]Z = -0.98[/tex] has a pvalue of 0.1635
16.35% probability that a randomly selected bag contains fewer than 1125 chocolate chips.
(c) What proportion of bags contains more than 1200 chocolate chips?
This is 1 subtracted by the pvalue of Z when X = 1200.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1200 - 1252}{129}[/tex]
[tex]Z = -0.40[/tex]
[tex]Z = -0.40[/tex] has a pvalue of 0.3446
0.3446 = 34.46% of bags contains more than 1200 chocolate chips
(d) What is the percentile rank of a bag that contains 1425 chocolate chips?
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1425 - 1252}{129}[/tex]
[tex]Z = 1.34[/tex]
[tex]Z = 1.34[/tex] has a pvalue of 0.91
So the answer is the 91st percentile.
