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

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