The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and standard deviation of 2 months. find the probability that an instrument produced by this machine will last q.1. between 7 and 12 months. q.2. less than 7 months q.3. more than 5 months but less than 10 months

Q.1. 0.4938

Q.2. 0.0062

Q.3. 0.1587

Step-by-step explanation:

Data

In the figure attached, the standard normal distribution table is shown.

Z = (x - μ)/σ

Q.1.

if x = 7, then Z = (7 - 12)/2 = -2.5

if x = 12, then Z = (12 - 12)/2 = 0

We want to find:

P(-2.5 < Z < 0) = P(Z < 0) - P(Z < -2.5) = 0.5 - 0.0062 = 0.4938

Q.2.

We want to find:

P(Z < -2.5) = 0.0062

Q.3.

if x = 5, then Z = (5 - 12)/2 = -3.5

if x = 10, then Z = (10 - 12)/2 = -1

We want to find:

P(-3.5 < Z < -1) = P(Z < -1) - P(Z < -3.5) = 0.1587 - 0 = 0.1587

answer: okw

step-by-step explanation:

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|