Question

The inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.)(a) Calculate P(12.99 ≤ X ≤ 13.01) when n = 16.P(12.99 ≤ X ≤ 13.01) =(b) How likely is it that the sample mean diameter exceeds 13.01 when n = 25?P(X ≥ 13.01) =

Answer 1

Answer:

a) P(12.99 ≤ X ≤ 13.01) = 0.3840

b) P(X ≥ 13.01) = 0.3075

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the cental 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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 13, \sigma = 0.08$

(a) Calculate P(12.99 ≤ X ≤ 13.01) when n = 16.

Here we have $n = 16, s = \frac{0.08}{\sqrt{16}} = 0.02$

This probability is the pvalue of Z when X = 13.01 subtracted by the pvalue of Z when X = 12.99.

X = 13.01

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

By the Central Limit Theorem

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

$Z = \frac{13.01 - 13}{0.02}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

X = 12.99

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

$Z = \frac{12.99 - 13}{0.02}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3075

0.6915 - 0.3075 = 0.3840

P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) How likely is it that the sample mean diameter exceeds 13.01 when n = 25?

P(X ≥ 13.01) =

This is 1 subtracted by the pvalue of Z when X = 13.01. So

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

$Z = \frac{13.01 - 13}{0.02}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

1 - 0.6915 = 0.3075

P(X ≥ 13.01) = 0.3075

Answer 2

Answer:

(a) P(12.99 ≤ X ≤ 13.01) = 0.3829

(b) P(X ≥ 13.01) = 0.2660

Step-by-step explanation:

We are given that the inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm. Suppose the distribution of the diameter is normal.

Firstly, using Central limit theorem the z score probability distribution is given by;

                Z = $\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = mean value = 13 cm

            $\sigma$ = standard deviation = 0.08 cm

            n = sample size = 16

(a) P(12.99 ≤ X ≤ 13.01) = P(X ≤ 13.01) - P(X < 12.99)

    P(X ≤ 13.01) = P( $\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ≤ $\frac{13.01-13}{\frac{0.08}{\sqrt{16} } }$ ) = P(Z ≤ 0.50) = 0.69146

    P(X < 12.99) = P( $\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{12.99-13}{\frac{0.08}{\sqrt{16} } }$ ) = P(Z < -0.50) = 1 - P(Z ≤ 0.50)

                                                          = 1 - 0.69146 = 0.30854

Therefore, P(12.99 ≤ X ≤ 13.01) = 0.69146 - 0.30854 = 0.3829

(b) Probability that sample mean diameter exceeds 13.01 when n = 25 is given by = P(X ≥ 13.01)

    P(X ≥ 13.01) = P( $\frac{X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ≥ $\frac{13.01-13}{\frac{0.08}{\sqrt{25} } }$ ) = P(Z ≥ 0.625) = 1 - P(Z < 0.625)

                                                         = 1 - 0.73401 = 0.2660

Hence, Probability that sample mean diameter exceeds 13.01 when n = 25 is 0.2660.