The inside diameter of a randomly selected piston ring is a random variable with mean value 9 cm and standard deviation 0.07 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.) Calculate P(8.99 < X < 9.01) when n = 16.
The inside diameter of a randomly selected piston ring is a random variable with mean value 9 cm and standard deviation 0.07 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.) Calculate P(8.99 < X < 9.01) when n = 16.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 22PFA
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![### Problem Statement: Understanding the Distribution of Piston Ring Diameters
**Description:**
The inside diameter of a randomly selected piston ring is a random variable with a mean value of 9 cm and a standard deviation of 0.07 cm. Suppose the distribution of the diameter is normal. The goal is to calculate the probability for a specific range of mean diameters when the sample size \( n = 16 \).
**Problem:**
Calculate \( P(8.99 \leq \bar{X} \leq 9.01) \) when \( n = 16 \).
**Instructions:**
Round your answers to four decimal places.
**Steps to Solve:**
1. Understand the problem is focused on a normally distributed variable, specifically the inside diameter of a piston ring.
2. Identify given parameters:
- Mean (\(\mu\)) = 9 cm
- Standard Deviation (\(\sigma\)) = 0.07 cm
- Sample size (\(n\)) = 16
3. Use these parameters to calculate the distribution of the sample mean \(\bar{X}\).
4. The standard error (SE) of the mean is calculated as:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{0.07}{\sqrt{16}} = \frac{0.07}{4} = 0.0175
\]
5. Convert the problem into the standard normal distribution (Z-distribution) using the Z-score formula:
\[
Z = \frac{\bar{X} - \mu}{SE}
\]
6. Calculate the Z-scores for the boundaries:
\[
Z_1 = \frac{8.99 - 9}{0.0175}, \quad Z_2 = \frac{9.01 - 9}{0.0175}
\]
7. Use the Z-scores to find the probabilities from the standard normal distribution tables or a calculator to get the final probability.
This problem aims to test the understanding of normal distribution, standard error, and Z-scores, which are fundamental concepts in statistics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb989860-4d1c-4fb5-9e6b-42a4528dce9c%2Fcd89f15b-9fd9-41c0-b642-ed5447378204%2Fu7x6xda_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement: Understanding the Distribution of Piston Ring Diameters
**Description:**
The inside diameter of a randomly selected piston ring is a random variable with a mean value of 9 cm and a standard deviation of 0.07 cm. Suppose the distribution of the diameter is normal. The goal is to calculate the probability for a specific range of mean diameters when the sample size \( n = 16 \).
**Problem:**
Calculate \( P(8.99 \leq \bar{X} \leq 9.01) \) when \( n = 16 \).
**Instructions:**
Round your answers to four decimal places.
**Steps to Solve:**
1. Understand the problem is focused on a normally distributed variable, specifically the inside diameter of a piston ring.
2. Identify given parameters:
- Mean (\(\mu\)) = 9 cm
- Standard Deviation (\(\sigma\)) = 0.07 cm
- Sample size (\(n\)) = 16
3. Use these parameters to calculate the distribution of the sample mean \(\bar{X}\).
4. The standard error (SE) of the mean is calculated as:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{0.07}{\sqrt{16}} = \frac{0.07}{4} = 0.0175
\]
5. Convert the problem into the standard normal distribution (Z-distribution) using the Z-score formula:
\[
Z = \frac{\bar{X} - \mu}{SE}
\]
6. Calculate the Z-scores for the boundaries:
\[
Z_1 = \frac{8.99 - 9}{0.0175}, \quad Z_2 = \frac{9.01 - 9}{0.0175}
\]
7. Use the Z-scores to find the probabilities from the standard normal distribution tables or a calculator to get the final probability.
This problem aims to test the understanding of normal distribution, standard error, and Z-scores, which are fundamental concepts in statistics.
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