6. The length of life of an instrument produced by a machine has a normal ditribution 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.

(a) .

(b) .

Step-by-step explanation:

We have been given that 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.

(a) First of all, we will find z-score corresponding to sample score of 7 months as:

, where,

z = Z-score,

x = Sample score,

= Mean,

= Standard deviation.    

Upon substituting our given values in z-score formula, we will get:

Now, we need to find the probability that a z-score is less than .

Using normal distribution table, we will get:

Therefore, the probability that an instrument produced by this machine will last less than 7 months is .

(b) Let us find z-score corresponding to sample score of 12 months.

Using formula , we will get:

Therefore, the probability that an instrument produced by this machine will last between 7 and 12 months is .

Step-by-step explanation:

Since the length of life of an instrument produced by a machine has a normal distribution, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = length of life of instruments in months.

µ = mean time

σ = standard deviation

From the information given,

µ = 12 months

σ = 2 months

a) We want to find the probability that an instrument produced by this machine will last for less than 7 months. It is expressed as

P(x < 7)

For x = 7,

z = (7 - 12)/2 = - 2.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.0062

b) between 7 and 12 months is expressed as P(7 ≤ x ≤ 12)

For x = 7, the probability is 0.0062

For x = 12,

z = (12 - 12)/2 = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

Therefore,

P(7 ≤ x ≤ 12) = 0.5 - 0.0062 = 0.4938

it's

your exact question is kind of difficult to understand, so maybe an example will explain what you're supposed to do? i'll work through the first problem on the worksheet.

suppose you have a tray of 652 brownies, and you have 5 friends that you want to feed without picking favorites. how many brownies can you dole out to each of your friends in equal amounts? how many will be left over for you to eat?

what you're doing here is finding a way to arrange the number 652 as the sum of integer multiple of 5 and some remainder:

here ("q" for quotient) is the number of brownies each of your friends get, and is the remainder. that's the basic idea behind division.

we break up the number 652 into parts so that the leading digit stands on its own:

652 = 600 + 52

first we check if it's possible to hand out exactly 100 brownies to each friend. with 5 friends, this means we want to know if we can take out 500 brownies from the tray. we can do this 1 time because we have 600 brownies to take from, and when we do that we're left with 100 + 52 brownies to divvy up. so we know that

652 = 1*500 + 152

next we want to know if it's possible to hand out exactly 10 brownies per friend. if we can, we have to remove 50 brownies from the tray each time until we can't. we have 152 brownies left, and we know that 3(50) = 150, so

152 = 3*50 + 2

now we have 2 brownies left in the tray. is it possible to hand 1 out to each of your 5 friends? no, you'd need a minimum of 5 brownies to do that. so the last 2 are yours.

putting everything together, we found that

652 = 1*500 + 3*50 + 2

and we can rearrange this to write

652 = 5*100 + 5*30 + 2

652 = 5(100 + 30) + 2

652 = 5*130 + 2

so and .

you might be more accustomed to the table depicting the algorithm. that's just a compact way of organizing what we've done above. you can consult the image i've attached below.