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Evgen [1.6K]
3 years ago
7

Jerry and Thomas own a cleaning business. Thomas cleans a house in 2 1/2 hours . Jerry cleans a house in 3 1/4 hours. How much l

onger does it take Jerry to clean a house than Thomas
Mathematics
1 answer:
allsm [11]3 years ago
7 0

Jerry takes 0.75 hours or \frac{6}{8} hours more to clean house than thomas

<h3><u>Solution:</u></h3>

Given that Jerry and Thomas own a cleaning business

Thomas cleans a house in 2\frac{1}{2} hours and Jerry cleans a house in 3\frac{1}{4} hours

To find: Number of hours more Jerry takes to clean than thomas

<em>Number of hours more Jerry takes to clean than thomas = time taken by Jerry - time taken by thomas</em>

Time taken by Jerry = 3\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{13}{4} \text{hours}

Time taken by Thomas = 2\frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2} \text{hours}

Substituting the values in above formula,

Number of hours more Jerry takes to clean than thomas = \frac{13}{4} - \frac{5}{2}

On solving we get,

=\frac{13}{4} - \frac{5}{2}\\\\= \frac{26-20}{8}\\\\= \frac{6}{8} = 0.75

Thus Jerry takes 0.75 hours or \frac{6}{8} hours more to clean house than thomas

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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:

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Central Limit Theorem

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

In this problem, we have that:

\mu = 21.3, \sigma = 3.9, n = 40, s = \frac{3.9}{\sqrt{40}} = 0.6166

Treating this as a random​ sample, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below?

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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X = 20.51

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