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The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 34,641 miles, with a standard deviation of 2480 miles. What is the probability that the sample mean would differ from the population mean by less than 344 miles in a sample of 121 tires if the manager is correct

User HBMCS
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2 Answers

6 votes

Answer:

87.27% probability that the sample mean would differ from the population mean by less than 344 miles in a sample of 121 tires if the manager is correct

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 = (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 = (\sigma)/(√(n)).

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:


\mu = 34631, \sigma = 2480, n = 121, s = (2480)/(√(121)) = 225.45

What is the probability that the sample mean would differ from the population mean by less than 344 miles in a sample of 121 tires if the manager is correct

This is the pvalue of Z when X = 34641 + 344 = 34985 subtracted by the pvalue of Z when X = 34641 - 344 = 34297. So

X = 34985


Z = (X - \mu)/(\sigma)

By the Central Limit Theorem


Z = (X - \mu)/(s)


Z = (34985 - 34631)/(225.45)


Z = 1.525


Z = 1.525 has a pvalue of 0.93635

X = 34297


Z = (X - \mu)/(s)


Z = (34297 - 34631)/(225.45)


Z = -1.525


Z = -1.525 has a pvalue of 0.06365

0.93635 - 0.06365 = 0.8727

87.27% probability that the sample mean would differ from the population mean by less than 344 miles in a sample of 121 tires if the manager is correct

User John Hedengren
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4.9k points
3 votes

Answer:


P(34641- 344 <\bar X < 34641 +344) = P(34297 <\bar X <34985)

And for this case we can use the z score formula given by:


z = (\bar X -\mu)/((\sigma)/(√(n)))

And if we use the formula for the limits that we have, we got:


z_1 = (34297-34641)/((2480)/(√(121)))= -1.526


z_2 = (34985-34641)/((2480)/(√(121)))= 1.526

So we want to find this probability and we can use the normal standard table and this difference:


P(-1.526 <Z<1.526) = P(z<1.526) -P(Z<-1.526) = 0.9365- 0.0635= 0.8730Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the following properties:

Where
\mu =34641 , \sigma =2480

Since the sample size used n =121 >30/ From the central limit theorem we know that the distribution for the sample mean
\bar X is given by:


\bar X \sim N(\mu, (\sigma)/(√(n)))

And we want to find this probability:


P(34641- 344 <\bar X < 34641 +344) = P(34297 <\bar X <34985)

And for this case we can use the z score formula given by:


z = (\bar X -\mu)/((\sigma)/(√(n)))

And if we use the formula for the limits that we have, we got:


z_1 = (34297-34641)/((2480)/(√(121)))= -1.526


z_2 = (34985-34641)/((2480)/(√(121)))= 1.526

So we want to find this probability and we can use the normal standard table and this difference:


P(-1.526 <Z<1.526) = P(z<1.526) -P(Z<-1.526) = 0.9365- 0.0635= 0.8730

User Aliza
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