Can someone see and answer this one
2 answers:
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Answer:
a) 30.08% probability the sample mean will be within $100 of the population mean.
b) 0% probability the sample mean will be greater than $12,600
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.
In this question, we have that:
a. What is the probability the sample mean will be within ±$100 of the population mean?
This is the pvalue of Z when X = 100 divided by s subtracted by the pvalue of Z when X = -100 divided by s. So
Z = 0.4 has a pvalue of 0.6554, Z = -0.4 has a pvalue of 0.3556
0.6554 - 0.3546 = 0.3008
30.08% probability the sample mean will be within $100 of the population mean.
b. What is the probability the sample mean will be greater than $12,600?
This is 1 subtracted by the pvalue of Z when X = 12600. So
By the Central Limit Theorem
has a pvalue of 1.
1 - 1 = 0
0% probability the sample mean will be greater than $12,600