Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a standard deviation of 1300 hours and a mean life span of 15,000 hours. If a monitor is selected at random, find the probability that the life span of the monitor will be more than 17,340 hours.

0.03593

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 17,340

μ is the population mean = 15,000

σ is the population standard deviation = 1300

z score:

z = 17340 - 15000/1300

z =1.8

Probability value from Z-Table:

P(x<17340) = 0.96407

P(x >17340) = 1 - P(x<17340)

= 1 - 0.96407

= 0.03593

The probability that the life span of the monitor will be more than 17,340 hours is 0.03593

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*