Respuesta :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 2.7, \sigma = 0.6[/tex].
(a) What is the probability that a randomly-selected customer will take less than 3 minutes?
This is the pvalue of Z when [tex]X = 3[/tex]. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3 - 2.7}{0.6}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] 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 [tex]s[/tex] instead of [tex]\sigma[/tex] in the formula of Z.
[tex]s = \frac{\sigma}{\sqrt{2}} = \frac{0.6}{\sqrt{2}} = 0.4243[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3 - 2.7}{0.4243}[/tex]
[tex]Z = 0.71[/tex]
[tex]Z = 0.71[/tex] 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.
(c) The probability that the average time of 64 randomly-selected customers will take less than 2.8 minutes is
This is the pvalue of Z when [tex]X = 2.8[/tex]
[tex]s = \frac{\sigma}{\sqrt{64}} = \frac{0.6}{8} = 0.075[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2.8 - 2.7}{0.075}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] 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.
[tex]s = \frac{\sigma}{\sqrt{81}} = \frac{0.6}{9} = 0.067[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2.8 - 2.7}{0.067}[/tex]
[tex]Z = 1.49[/tex]
[tex]Z = 1.49[/tex] 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.
[tex]s = \frac{\sigma}{\sqrt{225}} = \frac{0.6}{15}= 0.04[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2.8 - 2.7}{0.04}[/tex]
[tex]Z = 2.50[/tex]
[tex]Z = 2.50[/tex] 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.
[tex]s = \frac{\sigma}{\sqrt{400}} = \frac{0.6}{20}= 0.03[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2.8 - 2.7}{0.03}[/tex]
[tex]Z = 3.30[/tex]
[tex]Z = 3.30[/tex] 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.
