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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.

2 Answers

2 votes

Answer:

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

User Kaleigh
by
8.0k points
6 votes

Answer:

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

User Dimitri Bouniol
by
7.8k points

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