The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and standard deviation of 2 months. Find the probability that an instrument produced by this machine will last a) less than 7 months b) between 7 and 12 months.

Answer:a) 0.0062b) 0.4938Step-by-step explanation:a)We need to convert each to z score and use z-table to find the probabilities.The formula for z score is:[tex]z=\frac{x-\mu}{\sigma}[/tex]Where [tex]\mu[/tex] is the mean (given as 12), and[tex]\sigma[/tex]  is the standard deviation (given as 2)So we have:[tex]P(x<7)=P(z<\frac{7-12}{2})=P(z<-2.5)=0.0062[/tex]Hence, probability is 0.0062b)Here, we want between 7 and 12, we already found z-score of x = 7 to be -2.5. Let's find z score of x = 12 using the formula:[tex]z=\frac{x-\mu}{\sigma}\\z=\frac{12-12}{2}\\z=0[/tex]So we have:[tex]P(7<x<12)=P(-2.5<z<0) =0.4938[/tex]Hence, probability is 0.4938