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inysia [295]
2 years ago
15

The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of

a wedding is $29,858 (XO Group website, January 5, 2015). Assume that the cost of a wedding is normally distributed with a mean of $29,858 and a standard deviation of $5600.a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?
Mathematics
1 answer:
slavikrds [6]2 years ago
5 0

Answer:

a) 0.0392

b) 0.4688

c) At least $39,070 to be among the 5% most expensive.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

\mu = 29858, \sigma = 5600

a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?

This is the pvalue of Z when X = 20000. So

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

Z = \frac{20000 - 29858}{5600}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

So this probability is 0.0392.

b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?

This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.

X = 30000

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

Z = \frac{30000 - 29858}{5600}

Z = 0.02

Z = 0.02 has a pvalue of 0.5080.

X = 20000

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

Z = \frac{20000 - 29858}{5600}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

So this probability is 0.5080 - 0.0392 = 0.4688

c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?

This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.

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

1.645 = \frac{X - 29858}{5600}

X - 29858 = 5600*1.645

X = 39070

The wedding would have to cost at least $39,070 to be among the 5% most expensive.

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Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

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 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.

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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In this problem, we have that:

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This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

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

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

Z = \frac{3000 - 3500}{160}

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So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

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