The inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm

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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) =

Mathematics

2 answers:

Zina [9.1K] · Answer 1 · 2026-03-16T06:31:54+03:00

Answer:

(a) P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) P(X ≥ 13.01) = 0.3075

Step-by-step explanation:

To address this problem, one needs to grasp the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems involving normally distributed samples are resolved through the application of the z-score formula.

In a data set with a mean $\mu$ and standard deviation $\sigma$ , the z-score for a measurement X is defined as:

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

The Z-score indicates how many standard deviations the measurement deviates from the mean. After determining the Z-score, one can consult the z-score table to find the relevant p-value for that score. This p-value represents the probability that the measure's value is less than X, which indicates the percentile of X. By subtracting this p-value from 1, we obtain the likelihood that the measure's value exceeds X.

Central Limit Theorem

According to the Central Limit Theorem, 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 by a normal distribution, characterized by mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

The Central Limit Theorem can also be applicable to a skewed variable, provided that n is 30 or more.

In this case, we are given that:

$\mu = 13, \sigma = 0.08$

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

In this scenario, we find $n = 16, s = \frac{0.08}{\sqrt{16}} = 0.02$

This probability equates to the p-value of Z at X = 13.01 minus the p-value of Z at X = 12.99.

X = 13.01

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

According to the Central Limit Theorem

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

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

$Z = 0.5$

has a p-value of 0.6915

$Z = 0.5$

X = 12.99

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

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

$Z = -0.5$

$Z = -0.5$ yields a p-value of 0.3075

0.6915 - 0.3075 = 0.3840

P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) What is the probability that the sample mean diameter is greater than 13.01 when n = 25?

P(X ≥ 13.01) =

This is obtained by subtracting the p-value of Z when X = 13.01 from 1. Hence

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

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

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915

1 - 0.6915 = 0.3075

P(X ≥ 13.01) = 0.3075

Leona [9.2K] · Answer 2 · 2026-03-16T06:36:14+03:00

Answer:

(a) P(12.99 ≤ X ≤ 13.01) = 0.3829

(b) P(X ≥ 13.01) = 0.2660

Step-by-step explanation:

We know the inside diameter of a selected piston ring behaves as a random variable with a mean value of 13 cm and a standard deviation of 0.08 cm. Assuming a normal distribution for the diameter.

Initially, utilizing the Central Limit Theorem, the distribution of z-scores is described as;

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

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

(b) The likelihood of the sample mean diameter exceeding 13.01 when n = 25 is calculated as 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

Therefore, the probability that the sample mean diameter exceeds 13.01 when n = 25 is 0.2660.