Mathematics, 07.06.2020 00:01 XxMikeLearnsxX9288

Suppose a batch of metal shafts produced in a manufacturing company have a variance of 2.89 and a mean diameter of 211 inches. If 86 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.2 inches

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Answer from: nnn79

27.58% probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.2 inches

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(which is the square root of the variance) $\sigma$ , 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 $s = \frac{\sigma}{\sqrt{n}}$ .

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

In this question, we have that:

$\mu = 211, \sigma = \sqrt{2.89} = 1.7, n = 86, s = \frac{1.7}{\sqrt{86}} = 0.1833$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.2 inches.

Either greater than 211 + 0.2 = 211.2 or smaller than 211 - 0.2 = 210.8. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Probability of being less than 210.8:

This is the pvalue of Z when X = 210.8. So

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

By the Central Limit Theorem

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

$Z = \frac{210.8 - 211}{0.1833}$

Z = -1.09

Z = -1.09 has a pvalue of 0.1379

Probability of differing from the population mean by greater than 0.2 inches :

2*0.1379 = 0.2758

27.58% probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.2 inches

Answer from: Amholloway13

Probability is 86.24%

Step-by-step explanation:

We can solve this by using the Z-score formula.

Z = (X - μ)/σ

Where;

μ is mean = 211 inches

σ is standard deviation

Now, we are given variance as 2.89

Formula for standard deviation using variance is;

SD = √variance

SD = √2.89

SD = 1.7

To find the probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.2 inches, we will find the p-value of z and subtract 1 from it.

Since we are told that the sample shafts would differ by 0.2 inches, thus;

X = 211 + 0.2 = 211.2

Since we are working with sample mean of 86, then we have;

Z = (X - μ)/s

s = σ/√86

s = 1.7/√86

s = 0.1833

So,

Z = (211.2 - 211)/0.1833

Z = 1.0911

From z-score calculator, the p-value is gotten to be 0.1376

Thus, probability that mean diameter of the sample shafts would differ from the population mean by greater than 0.2 inches is;