Answer: 0.9306
Step-by-step explanation:
Given : A manufacturer knows that their items have a normally distributed lifespan, with a mean of 7.4 years, and standard deviation of 2.3 years.
i.e. [tex]\mu=7.4\ \ \&\ \sigma=2.3[/tex]
Let x denotes the lifespan of the items.
Then, the probability it will last longer than 4 years will be :-
[tex]P(x>4)=1-P(x\leq4)\\\\=1-P(\dfrac{x-\mu}{\sigma}<\dfrac{4-7.4}{2.3})\\\\=1-P(z<-1.48)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-(1-P(z<1.48))\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=P(z<1.48)= 0.9306\ \ [\text{By z-table}][/tex]
Hence , our required probability = 0.9306
