Can someone help..? thanks
1 answer:
You might be interested in
Answer:
The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, 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.
Mean $39725 and standard deviation $7320.
This means that
Sample of 125
This means that
The probability that the sample mean would be at least $39000 is about?
This is 1 subtracted by the pvalue of Z when X = 39000. So
By the Central Limit Theorem
has a pvalue of 0.1335
1 - 0.1335 = 0.8665
The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.