Suppose that the lifespan of laptops is normally distributed with a mean of 24.3 months and

standard deviation of 2.6 months. If USP provides its 33 staff with a laptop, find the probability

that the mean lifespan of these laptops will be less 23.8 months.

Step-by-step explanation:

Let x be the random variable representing the lifespan of laptops. Since the lifespans are normally distributed and population mean and population standard deviation are known, we would apply the formula,

z = (x - µ) / (σ/√n)

Where

x = sample mean

µ = population mean

σ = standard deviation

n = number of samples

From the information given,

µ = 24.5

σ = 2.6

n = 33

the probability that the mean lifespan of these laptops will be less 23.8 months is expressed as

P (x ≤ 23.8)

For x = 23.8

z = (23.8 - 24.5) / (2.6/√33) = - 1.55

Looking at the normal distribution table, the probability corresponding to area on the left of the z score is 0.061

P (x ≤ 23.8) = 0.061