Let's first find how many students bring their lunch to school. This is 30% of 300. To find this value, convert the percentage to a fraction (or decimal) and multiply:
Since 90 students bring their lunch to school, we can say 210 students do not bring their lunch to school. (300 - 90 = 210)
Answer:
a) By the Central Limit Theorem, the mean is $142 and the standard deviation is $0.7488.
b) By the Central Limit Theorem, approximately normal.
c) 0.0901 = 9.01% probability that the average cable service paid by the sample of cable service customers will exceed $143
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The mean monthly fee is $142 and the standard deviation is $29.
This means that
Part a: what are the mean an standard deviation of the sample distribution of x hat show your work and justify your reasoning.
Sample of 1500(larger than 30).
By the Central Limit Theorem
The mean is $142
The standard deviation is
Part b: what is the shape of the sampling distribution of x hat justify your answer.
By the Central Limit Theorem, approximately normal.
Part C: what is the probability that the average cable service paid by the sample of cable service customers will exceed $143?
This is 1 subtracted by the pvalue of Z when X = 143. So
By the Central Limit Theorem
has a pvalue of 0.9099
1 - 0.9099 = 0.0901
0.0901 = 9.01% probability that the average cable service paid by the sample of cable service customers will exceed $143