A university has 1,000 computers available for students to use. Each computer has a 250-gigabyte hard drive. The university wants to estimate the space occupied on the hard drives. A random sample of 100 computers showed a mean of 115 gigabytes used with a standard deviation of 20 gigabytes. What is the probability that a sample mean is between 111 and 119 gigabytes

0.9545

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/standard error,

where x is the raw score

μ is the population mean

Standard error = σ/√n

n = random number of samples

σ is the population standard deviation.

For x = 111

z = (x-μ)/standard error

= 111 - 115/20/√100

= -4 /20/10

= -4/2

= -2

Probalility value from Z-Table:

P(x≤ 111) = P(x = 111) = 0.02275

For x = 119

z = (x-μ)/standard error

= 119 - 115/20/√100

= 4 /20/10

= 4/2

= 2

Probability value from Z-Table:

P(x≤ 119) = P(x = 119) = 0.97725

The probability that a sample mean is between 111 and 119 gigabytes

P(-Z<x<Z)

= P(-2< x < 2)

= P(x = 119) - P(x = 111)

= 0.97725 - 0.02275

= 0.9545

i am thinking c. sorry if im wrong .

step-by-step explanation: