The time spent by a customer at a checkout counter has mean 4 minutes and standard deviation 2 minutes.
a. What is the probability that the total time taken by a random sample of 52 customers is less than 180 minutes?
b. Find the 30th percentile of the total time taken by 52 customers.

Respuesta :

The probability that the total time taken by a random sample of 52 customers is less than 180 minutes is 0.02%.

μ=Population mean=4

σ=Population standard deviation=2

n=Sample size=50

The sampling distribution of the sum S is roughly normal if the sample size is big (30 or more), according to the central limit theorem.

The central limit theorem tells us that the sampling distribution of the sum S is about normal because the sample size of 50 is at least 30.

The sum S's sample distribution has a mean and standard deviation of n and n, respectively.

The z-score is the value decreased by the mean, divided by the standard deviation

z = ( x -μS ) / √σS

= 150 - 50(4) / √ 50(2)

= 3.54

Using the appendix's normal probability table, get the relevant probability,

P(S < 150 ) = P(Z< -3.54)

= 0.0002

=0.02%

The term "standard deviation"  refers to a measurement of the data's dispersion from the mean. A low standard deviation implies that the data are grouped around the mean, whereas a large standard deviation shows that the data are more dispersed.

To learn more about standard deviation

brainly.com/question/23907081

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