Question

chapter 6 quiz - continuous probability distributions
10 points possible answered: 4/11
question 6
a company produces steel rods. the lengths of the steel rods are normally distributed with a mean of 153.5 - cm and a standard deviation of 0.6 - cm. for shipment, 43 steel rods are bundled together. round all answers to four decimal places if necessary.
a. what is the distribution of x? x ~ n( , )
b. what is the distribution of \\(\bar{x}\\)? \\(\bar{x}\\) ~ n( , )
c. for a single randomly selected steel rod, find the probability that the length is between 153.4 - cm and 153.5 - cm.
d. for a bundled of 43 rods, find the probability that the average length is between 153.4 - cm and 153.5 - cm.

Explanation:

Step1: Identify distribution of X

Given mean $\mu = 153.5$ and standard - deviation $\sigma=0.6$. For a single steel rod length $X$, the distribution is $X\sim N(153.5,0.6^{2})$.

Step2: Identify distribution of $\bar{X}$

The mean of the sampling distribution of the sample mean $\mu_{\bar{X}}=\mu = 153.5$, and the standard deviation of the sampling distribution of the sample mean (also known as the standard error) is $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$, where $n = 43$. So $\sigma_{\bar{X}}=\frac{0.6}{\sqrt{43}}\approx0.0912$. Then $\bar{X}\sim N(153.5,(\frac{0.6}{\sqrt{43}})^2)$.

Step3: Calculate $P(153.4\lt X\lt153.5)$ for a single rod

First, calculate the z - scores. The z - score formula is $z=\frac{x-\mu}{\sigma}$. For $x = 153.4$, $z_1=\frac{153.4 - 153.5}{0.6}\approx - 0.1667$. For $x = 153.5$, $z_2=\frac{153.5 - 153.5}{0.6}=0$. Then $P(153.4\lt X\lt153.5)=P(-0.1667\lt Z\lt0)=\varPhi(0)-\varPhi(-0.1667)$. Using the standard normal table, $\varPhi(0) = 0.5$ and $\varPhi(-0.1667)=1 - \varPhi(0.1667)\approx1 - 0.5662 = 0.4338$. So $P(153.4\lt X\lt153.5)=0.5 - 0.4338 = 0.0662$.

Step4: Calculate $P(153.4\lt\bar{X}\lt153.5)$ for a bundle of 43 rods

Calculate the z - scores for the sample mean. For $\bar{x}=153.4$, $z_1=\frac{153.4 - 153.5}{\frac{0.6}{\sqrt{43}}}\approx\frac{-0.1}{0.0912}\approx - 1.0965$. For $\bar{x}=153.5$, $z_2=\frac{153.5 - 153.5}{\frac{0.6}{\sqrt{43}}}=0$. Then $P(153.4\lt\bar{X}\lt153.5)=P(-1.0965\lt Z\lt0)=\varPhi(0)-\varPhi(-1.0965)$. Using the standard normal table, $\varPhi(-1.0965)=1 - \varPhi(1.0965)\approx1 - 0.8631 = 0.1369$. So $P(153.4\lt\bar{X}\lt153.5)=0.5 - 0.1369 = 0.3631$.

Answer:

a. $153.5,0.36$
b. $153.5,0.0083$
c. $0.0662$
d. $0.3631$