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2 years ago

11

A retail store did research and determined that, on average, each customer spends 40 minutes in their store per visit, with a st

andard deviation of 2 minutes. What percentage of customers spend more than 46 minutes?

1 answer:

Blababa [14]2 years ago

6 0

Answer:

0.13% of customers spend more than 46 minutes

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 40, \sigma = 2$

What percentage of customers spend more than 46 minutes?

This is 1 subtracted by the pvalue of Z when X = 46. So

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

$Z = \frac{46 - 40}{2}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% of customers spend more than 46 minutes

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

The formula to compute the future value is:

$FV=PV[1+\frac{r}{100}]^{n}$

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The expression to compute the amount in the investment account after 14 years is,

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Answer:

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Step-by-step explanation:

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3 years ago

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Zanzabum

Answer:

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Step-by-step explanation:

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We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

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Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

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$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

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3 years ago