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

Question

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?

in progress 0
Sadie 2 weeks 2021-11-24T04:42:39+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-11-24T04:44:33+00:00

    Answer:

    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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    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 \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

    In this problem, we have that:

    \mu = 7.65, \sigma = 2.15, n = 121, s = \frac{2.15}{\sqrt{121}} = 0.1955

    What is the probability that the sample average will be less than $8?

    This is the pvalue of Z when X = 8. So

    Z = \frac{X - \mu}{\sigma}

    By the Central Limit Theorem

    Z = \frac{X - \mu}{s}

    Z = \frac{8 - 7.65}{0.1955}

    Z = 1.79

    Z = 1.79 has a pvalue of 0.9633.

    96.33% probability that the sample average will be less than $8.

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )