A hardware store receives a shipment of bolts that are supposed to be 12 cm long. The mean is indeed 12 cm, and the standard deviation is 0.2 cm. For quality control, the hardware store chooses 100 bolts at random to measure. They will declare the shipment defective and return it to the manufacturer if the average length of the 100 bolts is less than 11.97 cm or greater than 12.04 cm. Find the probability that the shipment is found satisfactory.

Answer: 0.9104Step-by-step explanation:Given : A hardware store receives a shipment of bolts that are supposed to be 12 cm long. Mean : [tex]\mu=12\text { cm}[/tex]Standard deviation : [tex]\sigma= 0.2\text{ cm}[/tex]Sample size : n=10Since, they will declare the shipment defective and return it to the manufacturer if the average length of the 100 bolts is less than 11.97 cm or greater than 12.04 cm. So for the shipment to be satisfactory, the length of the bolts must be between 11.97 cm and 12.04 cm. We assume that the length of the bolts are normally distributed.Let X be the random variable that represents the length of randomly picked bolt .For Z score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]For x = 11.97 [tex]z=\dfrac{11.97-12}{\dfrac{0.2}{\sqrt{100}}}=-1.5[/tex]For x = 12.04[tex]z=\dfrac{12.04-12}{\dfrac{0.2}{\sqrt{100}}}=2[/tex]By using the standard normal distribution table ,  the probability that the shipment is found satisfactory will be :-[tex]P(511.97<X<12.04)=P(-1.5<z<2)=P(z<2)-P(z<-1.5)\\\\= 0.9772498-0.0668072=0.9104426\approx0.9104[/tex]Hence,  the probability that the shipment is found satisfactory=0.9104