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

Step-by-step 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) = $\frac{27-42}{9} = \frac{-15}{9}= \frac{-5}{3}=-1.666...$

and z (X₂ =39) = $\frac{39-42}{9} = \frac{-3}{9}= \frac{-1}{3}= -0.333...$

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%