Respuesta :
Answer:
Step-by-step explanation:
Given that X, the number of pizzas consumed per month by university students is normally distributed with a mean of 11 and a standard deviation of 3.
A) P(X>3) = P(Z>-2.33) = 0.9901
B) n=11 hence std error = [tex]\frac{3}{\sqrt{9} } =1[/tex]
Each student eats more than 132/11=12 pizzas
P(x>12) = P(Z>3) = 0.5-0.4987=0.0013
Using the normal probability distribution and the central limit theorem, it is found that there is a:
a) 0.3707 = 37.07% students consume more than 12 pizzas per month.
b) 0.1335 = 13.35% probability that in a random sample of size 11, a total of more than 132 pizzas are consumed.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the standard deviation of the sampling distribution of samples of size n is given by [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem:
- Mean of 11, thus [tex]\mu = 11[/tex]
- Standard deviation of 3, thus [tex]\sigma = 3[/tex].
Item a:
This proportion is 1 subtracted by the p-value of Z when X = 12, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12 - 11}{3}[/tex]
[tex]Z = 0.33[/tex]
[tex]Z = 0.33[/tex] has a p-value of 0.6293.
1 - 0.6293 = 0.3707.
0.3707 = 37.07% students consume more than 12 pizzas per month.
Item b:
- Now sample of size 11, thus [tex]n = 11, s = \frac{3}{\sqrt{11}}[/tex].
- Sum of 132 is a sample mean of [tex]\frac{132}{11} = 12[/tex], thus, the probability is 1 subtracted by the p-value of Z when X = 12.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{12 - 11}{\frac{3}{\sqrt{11}}[/tex]
[tex]Z = 1.11[/tex]
[tex]Z = 1.11[/tex] has a p-value of 0.8665.
1 - 0.8665 = 0.1335
0.1335 = 13.35% probability that in a random sample of size 11, a total of more than 132 pizzas are consumed.
A similar problem is given at https://brainly.com/question/24663213
