Question

Suppose a batch of metal shafts produced in a manufacturing company have a variance of 7.84 and a mean diameter of 201 inches. If 89 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would be greater than 201.2 inches? Round your answer to four decimal places.

Answer 1

Answer:

There is a 25.14% probability that the mean diameter of the sample shafts would be greater than 201.2 inches

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

Suppose a batch of metal shafts produced in a manufacturing company have a variance of 7.84 and a mean diameter of 201 inches. This means that $\mu = 201$.

The standard deviation is the square root of the variance. So $\sigma = \sqrt{7.84} = 2.8$

If 89 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would be greater than 201.2 inches?

This 1 subtracted by the pvalue of Z when $X = 201.2$

We also have to find a standard deviation of the sample(that is, the 89 shafts), as we are working with the sample's mean, so:

$s = \frac{\sigma}{\sqrt{89}} = 0.2968$

So

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

$Z = \frac{201.2-201}{0.2968}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

This means that there is a 1-0.7486 = 0.2514 = 25.14% probability that the mean diameter of the sample shafts would be greater than 201.2 inches

Answer 2

The probability that the mean diameter of the sample shafts would be greater than 201.2 inches is 0.2514 approx.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

$X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$ )

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

$P(Z \leq z) = P(Z < z) )$

Also, know that if we look for Z = z in z tables, the p value we get is

$P(Z \leq z) = \rm p \: value$

For this case, we have:

• Population mean = $\mu = 201 \: \rm in.$
• Population variance = $\sigma^2= 7.84 \: \rm in^2$
• Population standard deviation $\sigma = \sqrt{7.84} = 2.8 \: \rm in$
• Sample size = n = 89

(it is because standard deviation is positive sq. root of variance)

If we take X = diameter of shafts in the population, then:

$X \sim N(\mu =201, \sigma =2.8)$

Let Y be diameter of shafts in the  selected sample, then:

$Y \sim N(\overline{x}, s)$

where $\overline{x}$ = sample mean, and s= sample standard deviation.

Since sample size is large enough(89 > 30), we can take $\overline{x} = \mu = 201$ and $s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{89}} \approx 0.297$

$P(Y > 201.2) = P(Z > z \approx \dfrac{201.2 - 201}{0.297}) = P(Z > 0.673)\\\\P(Y > 201.2) = 1 - P(Z \leq 0.673)$

where Z denotes the standard normal distribution.

Using the z-tables, the p-value for Z = 0.67 is obtained

Thus, we get:

$P(Y > 201.2) \approx 1 - P(Z \leq 0.673) \approx 1 -0.7486= 0.2514$

Thus, the probability that the mean diameter of the sample shafts would be greater than 201.2 inches is 0.2514 approx.

Learn more about standard normal distribution here:

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