Answer :
Answer:
There is a 7.18% probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1.
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:
At a large university, the mean amount spent by students for cell phone service is $38.90 per month with a standard deviation of $3.64 per month.
This means that [tex]\mu = 38.90, \sigma = 3.64[/tex]
Consider a group of 44 randomly chosen university students. What is the probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1?
By the Central Limit Theorem, this means that now we have to use the standard deviation of the mean, since we are working with a sampling distribution of the sample mean. So:
[tex]s = \fra{\sigma}{\sqrt{44}} = \frac{3.68}{\sqrt{44}} = 0.5548[/tex]
This probability is:
[tex]P = P(X \geq 39.90) + P(X \leq 37.90)[/tex]
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[tex]P = P(X \geq 39.90)[/tex]
This probability is 1 subtracted by the pvalue of Z when [tex]X = 39.90[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{39.90 - 38.90}{0.5548}[/tex]
[tex]Z = 1.80[/tex]
[tex]Z = 1.80[/tex] has a pvalue of 0.9641. This means that [tex]P(X \geq 39.90) = 1-0.9641 = 0.0359[/tex].
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[tex]P = P(X \leq 37.90)[/tex]
This probability is the pvalue of Z when [tex]Z = 37.90[/tex].
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{37.90 - 38.90}{0.5548}[/tex]
[tex]Z = -1.80[/tex]
[tex]Z = -1.80[/tex] has a pvalue of 0.0359. So [tex]P(X \leq 37.90) = 0.0359[/tex].
So
[tex]P = P(X \geq 39.90) + P(X \leq 37.90) = 2*0.0359 = 0.0718[/tex]
There is a 7.18% probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1.
The probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1 is 0.0688.
How to calculate probability?
From the information given, the mean is $38.90 per month, standard deviation of $3.64 per month, and the sample is 44.
The sampling distribution of the sampling mean will be:
= 3.64/✓44 = 0.5494
The probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1 will be:
= 1/0.5494
= 1.82
Therefore, P(z > 1.82) from the z table will be 0.0688.
In conclusion, the probability is 0.0688.
Learn more about probability on:
https://brainly.com/question/24756209