# A machine produces parts with lengths that are normally distributed with displaystylesigma={0.66}. A sample of 20 parts has a mean length of 75.98. a)

A machine produces parts with lengths that are normally distributed with $\sigma =0.66$. A sample of 20 parts has a mean length of 75.98.
a)
Give a point estimate for $\mu$. (Give your answer correct to two decimal places.)
b)
Find the $95\mathrm{%}$ confidence maximum error of estimate for $\mu$. (Give your answer correct to two decimal places.)
c)
Find the $95\mathrm{%}$ confidence interval for $\mu$. (Give your answer correct to two decimal places.)
Lower limit $=?$
Upper limit $=?$
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Step 1
Given:
A machine produces parts with lengths that are normally distributed with $\sigma =0.66$. A sample of 20 parts has a mean length of 75.98.
To Determine:
a) A point estimate for $\mu .$
b) The $95\mathrm{%}$ confidence maximum error of estimate for $\mu .$
c) The $95\mathrm{%}$ confidence interval for $\mu$
Step 2
Solution:
A machine produces parts with lengths that are normally distributed with the given values,
$\sigma =0.66,n=20,\stackrel{―}{x}=75.98$
a) A point estimate for $\mu$ is
$\stackrel{―}{\mu }=\stackrel{―}{x}=75.98$
b) For $95\mathrm{%}$ confidence intervals,
Z - critical value is 1.96.
So, The maximum confidence error of estimate for $\mu$ is,
$M.E.=\frac{Z×\sigma }{\sqrt{n}}$
$M.E.=\frac{1.96×0.66}{\sqrt{20}}$
$M.E.=\frac{1.96×0.66}{4.4721}$
$M.E.=\frac{1.2936}{4.4721}=0.289260$
c) The $95\mathrm{%}$ confidence interval for $\mu$,
Lower Limit $=\stackrel{―}{x}-M.E.$
L.L.,
$=75.98-0.289260$
$=75.69074$
Upper Limit $=\stackrel{―}{x}+M.E.$
U.L.,
$=75.98+0.289260$
$=76.26926$