Respuesta :
The probability of a randomly selected hard drive from the company lasting between 2 years 3 months and 3 years 3 months is 32.22%
Given here,
Mean (μ) = 3 years 6 months
= (3×12)+6 = 42 months
Standard deviation (σ) = 9 months
We will find the z-score using the formula: z = (X - μ)/σ
Here X₁ = 2 years 3 months
= (2×12)+3 = 27 months
and X₂ = 3 years 3 months
= (3×12)+3 = 39 months
So, z (X₁ =27) = [tex] \frac{27-42}{9} = \frac{-15}{9}= \frac{-5}{3}=-1.666... [/tex]
and z (X₂ =39) = [tex] \frac{39-42}{9} = \frac{-3}{9}= \frac{-1}{3}= -0.333... [/tex]
According to the standard normal table,
P(z> -1.666...) = 0.0485 and P(z< -0.333...) = 0.3707
So, P(27 < X < 39)
= 0.3707 - 0.0485
= 0.3222
= 32.22 % [Multiplying by 100 for getting percentage]
So, the probability of a randomly selected hard drive from the company lasting between 2 years 3 months and 3 years 3 months is 32.22%
