Question

The life spans of a computer manufacturer’s hard drives are normally distributed, with a mean of 3 years 6 months and a standard deviation of 9 months. What is the probability of a randomly selected hard drive from the company lasting between 2 years 3 months and 3 years 3 months? Use the portion of the standard normal table below to help answer the question.

Answer 1

Answer: A on edg

Explanation:

Answer 2

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) =
`(27-42)/(9) = (-15)/(9)= (-5)/(3)=-1.666...`

and z (X₂ =39) =
`(39-42)/(9) = (-3)/(9)= (-1)/(3)= -0.333...`

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%