Mathematics, 21.04.2020 04:13, ishmael9332

The shape of the distribution of the time required to get an oil change at a 20 minute oil change facility. However, records indicate the mean time is 21.3 minutes and the standard deviation is 3.9 minutes.

Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 A. M. and 12 P. M. Treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the manager.

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answered: kkkhomewirk

$20.51 \rm min$ mean oil-change time would there be a 10% chance of being at or below.

Step-by-step explanation:

Given information:

Mean time $\mu=21.3\rm min$

Standard deviation $\sigma=3.9\rm min$

n=40

standard error $s=\frac{\sigma}{\sqrt n}=\frac{3.9}{40}=0.6166$

10% chance of being at or below, $\widehat{p}=0.1$

By use of standard normal table

$\widehat{p}=0.1$ , z=-1.282

By Central limit theorem,

$z=\frac{X-\mu}{s}$

On substitution,

$-1.282=\frac{X-21.3}{\ 0.6166}$

$X=20.51 \rm min$

Hence, $20.51 \rm min$ mean oil-change time would there be a 10% chance of being at or below.

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answered: dsteele32

For a mean oil change time of 20.51 minutes there would be a 10% chance of being at or below

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .