Question

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

Answer 1

Q.1. 0.4938

Q.2. 0.0062

Q.3. 0.1587

Explanation:

Data

• mean, μ = 12

• standard deviation, σ = 2

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 2

solved

Explanation:

Mean = 12

SD = 2

Answer to part 2) Less than 7 months

P(x < 7)

P(x < 7) = P(z < (7-12)/2) = P(z < -2.5)

This probability can be obtained by referring to the Z table

Thus P(x < 7) = 0.0062

1) Between 7 and 12 months

P(7 < x < 12) = P( x< 12) - P(x < 7)

We have P(x < 12) = 0 ......[because mean = 12]

and P(x < 7) = 0.0062 ...[this we got is part a]

Thus on plugging these values we get

P( 7 < x < 12) = 0.5 - 0.0062 = 0.4938

Thus P(7 < x < 12) = 0.4938

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