Mathematics, 10.11.2019 05:31 lapointayyy6388

The mean checkout time in the express lane of a large grocery store is 2.7 minutes, and the standard deviation is 0.6 minutes. the distribution of checkout times is non-normal (for one thing, it can be a lot longer than 2.7 minutes, but it can only be so short).
(a) what is the probability that a randomly-selected customer will take less than 3 minutes? 0.3085 0.6915 it cannot be determined from the information given.
(b) what is the probability that the average time of two randomly-selected customers will take less than 3 minutes? 0.7611 0.2389 it cannot be determined from the information given. a
(c) the probability that the average time of 64 randomly-selected customers will take less than 2.8 minutes is.
(d) the probability that the average time of 81 randomly-selected customers will take less than 2.8 minutes is.
(e) the probability that the average time of 225 randomly-selected customers will take less than 2.8 minutes is.
(f) the probability that the average time of 400 randomly-selected customers will take less than 2.8 minutes is.

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Answer from: kelechikamalu1

a) There is a 69.155 probability that a randomly-selected customer will take less than 3 minutes.

b) There is a 76.11% probability that the average time of two randomly-selected customers will take less than 3 minutes.

c) There is a 90.82% probability that the average time of 64 randomly-selected customers will take less than 2.8 minute.

d) There is a 93.19% probability that the average time of 81 randomly-selected customers will take less than 2.8 minutes.

e) There is a 99.38% probability that the average time of 225 randomly-selected customers will take less than 2.8 minutes.

f) There is a 99.95% probability that the average time of 400 randomly-selected customers will take less than 2.8 minutes.

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:

The mean checkout time in the express lane of a large grocery store is 2.7 minutes, and the standard deviation is 0.6 minutes. This means that $\mu = 2.7, \sigma = 0.6$ .

(a) What is the probability that a randomly-selected customer will take less than 3 minutes?

This is the pvalue of Z when X = 3 . So:

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

$Z = \frac{3 - 2.7}{0.6}$

Z = 0.5

Z = 0.5 has a pvalue of 0.6915.

There is a 69.155 probability that a randomly-selected customer will take less than 3 minutes.

(b) What is the probability that the average time of two randomly-selected customers will take less than 3 minutes?

We are now working with the average of a sample, so we must use instead of $\sigma$ in the formula of Z.

$s = \frac{\sigma}{\sqrt{2}} = \frac{0.6}{\sqrt{2}} = 0.4243$

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

$Z = \frac{3 - 2.7}{0.4243}$

Z = 0.71

Z = 0.71 has a pvalue f 0.7611.

There is a 76.11% probability that the average time of two randomly-selected customers will take less than 3 minutes.

This is the pvalue of Z when X = 2.8

$s = \frac{\sigma}{\sqrt{64}} = \frac{0.6}{8} = 0.075$

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

$Z = \frac{2.8 - 2.7}{0.075}$

Z = 1.33

Z = 1.33 has a pvalue of 0.9082.

There is a 90.82% probability that the average time of 64 randomly-selected customers will take less than 2.8 minute.

(d) The probability that the average time of 81 randomly-selected customers will take less than 2.8 minutes is.

$s = \frac{\sigma}{\sqrt{81}} = \frac{0.6}{9} = 0.067$

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

$Z = \frac{2.8 - 2.7}{0.067}$

Z = 1.49

Z = 1.49 has a pvalue of 0.9319.

There is a 93.19% probability that the average time of 81 randomly-selected customers will take less than 2.8 minutes.

(e) The probability that the average time of 225 randomly-selected customers will take less than 2.8 minutes is.

$s = \frac{\sigma}{\sqrt{225}} = \frac{0.6}{15}= 0.04$

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

$Z = \frac{2.8 - 2.7}{0.04}$

Z = 2.50

Z = 2.50 has a pvalue of 0.9938.

There is a 99.38% probability that the average time of 225 randomly-selected customers will take less than 2.8 minutes.

(f) The probability that the average time of 400 randomly-selected customers will take less than 2.8 minutes is.

$s = \frac{\sigma}{\sqrt{400}} = \frac{0.6}{20}= 0.03$

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

$Z = \frac{2.8 - 2.7}{0.03}$

Z = 3.30

Z = 3.30 has a pvalue of 0.9995.

There is a 99.95% probability that the average time of 400 randomly-selected customers will take less than 2.8 minutes.

Answer from: Quest

imma say 16%. sorry if its wrong. i'm doing the same stuff right now

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

i'm pretty sure it would take 20 hours because it's half the men so you multiply the number of hours it takes by 2

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