The mean amount purchased by a typical customer at Churchill’s Grocery Store is $27.50 with a standard deviation of $7.00. Assume the distri

Question

The mean amount purchased by a typical customer at Churchill’s Grocery Store is $27.50 with a standard deviation of $7.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 68 customers, answer the following questions

a. What is the likelihood the sample mean is at least $30.00?
b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?
c. Within what limits will 90 percent of the sample means occur?

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Thạch Thảo 5 years 2021-08-18T03:51:16+00:00 1 Answers 96 views 0

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  1. kiet
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    2021-08-18T03:52:45+00:00
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    Answer:

    a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.

    b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00

    c) 90% of sample means will occur between $26.1 and $28.9.

    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 \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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

    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:

    \mu = 27.50, \sigma = 7, n = 68, s = \frac{7}{\sqrt{68}} = 0.85

    a. What is the likelihood the sample mean is at least $30.00?

    This is 1 subtracted by the pvalue of Z when X = 30. So

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

    By the Central Limit Theorem, we have that:

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

    Z = \frac{30 - 27.5}{0.85}

    Z = 2.94

    Z = 2.94 has a pvalue of 0.9984

    1 – 0.9984 = 0.0016

    0.0016 = 0.16% probability that the sample mean is at least $30.00.

    b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?

    This is the pvalue of Z when X = 30 subtracted by the pvalue of Z when X = 26.50. So

    From a, when X = 30, Z has a pvalue of 0.9984

    When X = 26.5

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

    Z = \frac{26.5 - 27.5}{0.85}

    Z = -1.18

    Z = -1.18 has a pvalue of 0.1190

    0.9984 – 0.1190 = 0.8794

    0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00.

    c. Within what limits will 90 percent of the sample means occur?

    Between the 50 – (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile, that is, Z between -1.645 and Z = 1.645

    Lower bound:

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

    -1.645 = \frac{X - 27.5}{0.85}

    X - 27.5 = -1.645*0.85

    X = 26.1

    Upper Bound:

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

    1.645 = \frac{X - 27.5}{0.85}

    X - 27.5 = 1.645*0.85

    X = 28.9

    90% of sample means will occur between $26.1 and $28.9.

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