The distribution of the time that it takes a group of mechanics to do an oil change at a garage shop is unknown. However, it takes the mechanics an average of μ = 20 minutes to do an oil change with a standard deviation of σ = 2 minutes. A manager at this shop draws a sample of 36 oil changes to monitor the mechanics' efficiency. Let X represent the sample mean time for these 36 oil changes. Select all that apply.

Step-by-step explanation:

Hello!

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You have the variable of interest:

X: the time it takes a mechanic to do an oil change at a garage shop.

This variable has an unknown distribution but the population mean is known to be μ= 20 min and the population standard deviation δ= 2 min (then the variance is δ²= 4min)

Suppose that you need to study the average time it takes the mechanics to do an oil change, but you don't have any idea of the shape of this population's distribution, then you can apply the Central Limit Theorem to approximate the sampling distribution to normal.

This theorem states that:

Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

X[bar]≈N(μ;σ²/n)

Then since n=36 oil changes, you can approximate the sampling distribution to normal:

Mean: μ= 20 min

Variance: σ²/n= 4/36= 1/9= 0.11min

X[bar]≈N(20;0.11)

Using this approximation you can study the population mean (trough Hypothesis test or estimation) or calculate the probabilities for given values of the sample mean.

I hope this helps a little!

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step-by-step explanation:

y -5 = (-1/4)(x + 8)

step-by-step explanation: