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simiyi1983
24.02.2020 •
Mathematics
A company is interested in determining the average cost of lunch in its cafeteria, so it asks a sample of 121 randomly chosen employees what they spent. If the true average amount spent on lunch is $7.65 with a standard deviation of $2.15, what is the probability that the sample average will be less than $8?
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Ответ:
96.33% probability that the sample average will be less than $8.
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using 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 random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
In this problem, we have that:
What is the probability that the sample average will be less than $8?
This is the pvalue of Z when X = 8. So
By the Central Limit Theorem
96.33% probability that the sample average will be less than $8.
Ответ: