An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with a mean

An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with a mean of 116 cm and a standard deviation of 4.8 cm.
A. Find the probability that one selected subcomponent is longer than 118 cm.
B. Find the probability that if 4 subcomponents are randomly selected, their mean length exceeds 118 cm.
C. Find the probability that if 4 are randomly selected, all 4 have lengths that exceed 118 cm.
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Step 1
Given Data:
$X\sim N\left(116,{4.8}^{2}\right)$
A) To Find: $P\left(X>118\right)$
$P\left(X>118\right)=P\left(X-\frac{\mu }{\sigma }>118-\frac{\mu }{\sigma }\right)$
$=P\left(z>\frac{118-116}{4.8}\right)$
$=P\left(z>0.4167\right)$
$=1-P\left(z<0.4167\right)$
$=1-0.662=0.338$ (using s tan dard normal distribution table)
The probability that one selected subcomponent is longer than 118 cm is 0.338
Step 2
B) sample size=4
Therefore,$\stackrel{―}{X}\sim N\left(116,\frac{{4.8}^{2}}{4}\right)$
To Find:$P\left(\stackrel{―}{X}>118\right)$
$P\left(\stackrel{―}{X}>118\right)=P\left(\stackrel{―}{X}-\frac{\mu }{\sigma }/\sqrt{n}>118-\frac{\mu }{\sigma }/\sqrt{n}\right)$
$=P\left(z>118-\frac{116}{4.8}/\sqrt{4}\right)$
$=P\left(z>0.8333\right)$
$=1-P\left(z<0.8333\right)$
$=1-0.798=0.202$ (using standard normal distrbution table)
The probability that if 4 subcomponents are randomly selected, their mean length exceeds 118 cm is 0.202
Step 3 C) P(all 4 lengths exceed 118 cm)=$P\left(X>118{\right)}^{4}$
$={0.338}^{4}$ (From part A, $P\left(X>118\right)=0.338\right)=0.0131$
The probability that if 4 are randomly selected, all 4 have lengths that exceed 118 cm is 0.0131