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. z probability 0.00 0.5000 0.23 0.5910 0.33 0.6293 0.67 0.7486 1.00 0.8413 1.33 0.9082 1.67 0.9525 2.00 0.9772 32% 37% 42% 95%

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2011553 2011553 · Answer 1 · 2022-03-26T00:33:12+01:00

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) =

and z (X₂ =39) =

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%