Question

Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 3,240,0003,240,000 and a mean life span of 17,00017,000 hours. If a monitor is selected at random, find the probability that the life span of the monitor will be more than 19,14119,141 hours. Round your answer to four decimal places.

Answer 1

Answer:

0.1172 is the probability that the life span of the monitor will be more than 19,141 hours.

Explanation:

We are given the following information in the question:

Mean, μ = 17,000

Variance = 3,240,000

Standard deviation, σ = 1800

We are given that the distribution of life spans is a bell shaped distribution that is a normal distribution.

Formula:

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

P(life span of the monitor will be more than 19,141)

P(x > 19141)

$P( x > 19141) = P( z > \displaystyle\frac{19141 - 17000}{1800}) = P(z > 1.1891)$

$= 1 - P(z \leq 1.1891)$

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

$P(x > 19141) = 1 - 0.8828 = 0.1172=11.72\%$

0.1172 is the probability that the life span of the monitor will be more than 19,141 hours.