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 a) less than 7 months b) between 7 and 12 months.

Answer 1

Explanation:

a)

We need to convert each to z score and use z-table to find the probabilities.

The formula for z score is:

Where is the mean (given as 12), and

is the standard deviation (given as 2)

So we have:

Hence, probability is 0.0062

b)

Here, we want between 7 and 12, we already found z-score of x = 7 to be -2.5. Let's find z score of x = 12 using the formula:

So we have:

Hence, probability is 0.4938

Answer 2

a) 0.0062

b) 0.4938

Explanation:

a)

We need to convert each to z score and use z-table to find the probabilities.

The formula for z score is:

`z=(x-\mu)/(\sigma)`

Where
`\mu` is the mean (given as 12), and

`\sigma` is the standard deviation (given as 2)

So we have:

`P(x<7)=P(z<(7-12)/(2))=P(z<-2.5)=0.0062`

Hence, probability is 0.0062

b)

Here, we want between 7 and 12, we already found z-score of x = 7 to be -2.5. Let's find z score of x = 12 using the formula:

`z=(x-\mu)/(\sigma)\\z=(12-12)/(2)\\z=0`

So we have:

`P(7<x<12)=P(-2.5<z<0) =0.4938`

Hence, probability is 0.4938