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lakkis [162]
3 years ago
5

Suppose the waiting times in an emergency room are normally distributed with a mean (mean) of 40 and a standard deviation (sigma

) of 10. A random sample of 25 persons waiting in the ER is selected. What is the probability that their average waiting time will be between 38 and 42 minutes
Mathematics
1 answer:
kipiarov [429]3 years ago
3 0

Answer:

0.68269

Step-by-step explanation:

When we are to find the z score for population where a random sample is picked, the z.score formula we use is

z = (x-μ)/Standard error, where

x is the raw score,

μ is the population mean

Standard error = σ/√n

σ is the population standard deviation

n = random number of samples

For : x = 38 minutes, μ = 40, σ = 10, n = 5

z = 38 - 40/10 /√25

= -2/10/5

= -2/2

= -1

Determining the probability value using z table

P(x = 38) = P(z = -1)

= 0.15866

For : x = 42 minutes, μ = 40, σ = 10, n = 25

z = 42 - 40/10 /√25

= 2/10/5

= 2/2

= 1

Determining the probability value using z table

P(x = 42) = P(z = 1)

= 0.84134

The probability that their average waiting time will be between 38 and 42 minutes is calculated as

P(-Z<x<Z)

= P(-1 < x < 1)

= P(z = 1) - P(z = -1)

= 0.84134 - 0.15866

= 0.68269

Therefore, the probability that their average waiting time will be between 38 and 42 minutes is 0.68269

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