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Answer:
0.8665 = 86.65% probability that the sample mean would be at least $39000
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
0.8665 = 86.65% probability that the sample mean would be at least $39000
well, we know the common difference is -3, to go from the 4th term to the 8th term, we need to add "d" 4 times or namely 3+4(-3), likewise to go from the 13th term to the 19th term we have to add "d" 6 times, or namely -24 + 6(-3).