A manufacturer knows that their items have a normally distributed lifespan, with a mean of 12.3 years, and standard deviation of 0.8 years. If you randomly purchase one item, what is the probability it will last longer than 14 years

probability of lasting longer = 1.7%

Step-by-step explanation:

We are given:

x' = 14 years

μ = 12.3 years

s = 0.8 years

Thus, let's use the formula for the Z-score value which is;

z = (x' - μ)/s

Thus;

z = (14 - 12.3)/0.8

z = 2.125

From the z-distribution table attached, the p-value is ;

P(x' > 2.125) = 1 - 0.983 = 0.017 = 1.7%

Thus,probability of lasting longer = 1.7%