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The quality control manager of at Long John Silver's restaurant wants to analyze the length of time that a car spends at the drive-thru window waiting for an order. It is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1 seconds. The distribution of time at the window is skewed right. The quality control manager wishes to use a new delivery system designed to get cars through the drive-thru system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. What is the probability of obtaining a sample mean of 56.8 seconds or less, assuming that the population mean is 59.3 seconds

Respuesta :

Answer:

0.11314

Step-by-step explanation:

The Long John Silver's restaurant parameters are;

The mean time spent at the window, μ = 59.3

The standard deviation, σ = 13.1

The time distribution is skewed right

The number of cars in the sample, n = 40 cars

The mean time spent at the window with the delivery system, [tex]\overline x[/tex] = 56.8 seconds

The Z-value for a mean time of 56.8 seconds is given as follows;

[tex]Z=\dfrac{\bar{x}-\mu }{\dfrac{\sigma }{\sqrt{n}}}[/tex]

Therefore, we get;

[tex]Z=\dfrac{56.8-59.3 }{\dfrac{13.1 }{\sqrt{40}}} \approx -1.20697620617[/tex]

Therefore, the z-value ≈ -1.21

From the normal probability table, we have;

P([tex]\overline X[/tex] ≤ 56.8) = P(Z < -1.21) = 0.11314

The probability of obtaining a sample mean of 56.8 second or less, assuming that the population mean is 59.3 seconds P([tex]\overline X[/tex] ≤ 56.8) = 0.11314.

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