Question

The time taken for a computer to boot up, X, follows a normal distribution with mean 30 seconds and standard deviation 5 seconds. What is the probability that a computer will take more than 42 seconds to boot up?

Answer 1

Answer:

0.008 is the probability that a computer will take more than 42 seconds to boot up.                                  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 30 seconds

Standard Deviation, σ = 5 second

We are given that the distribution of time taken for a computer to boot up is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(computer will take more than 42 seconds to boot up)

P(x > 42)

$P( x > 42) = P( z > \displaystyle\frac{42 - 30}{5}) = P(z > 2.4)$

$= 1 - P(z \leq 2.4)$

Calculation the value from standard normal z table, we have,  

$P(x > 42) = 1 - 0.992 = 0.008$

0.008 is the probability that a computer will take more than 42 seconds to boot up.