We are given that the population mean for the battery life is `mu=22.5` with a population standard deviation of `sigma=1.58`, and that the population is approximately normal .

(1) Find the probability that a pair of batteries lasts between 20 and 24 hours, or `P(20<X<24)` :

To find the probability...

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We are given that the population mean for the battery life is `mu=22.5` with a population standard deviation of `sigma=1.58`, and that the population is approximately normal .

(1) Find the probability that a pair of batteries lasts between 20 and 24 hours, or `P(20<X<24)` :

To find the probability that a single data point lies on an interval, we convert to a standard score and solve the resulting probability question. Since we know the population standard deviation, we use the `z` table (standard normal table).

To normalize a data point we use `z=(X-mu)/sigma` so:

`z_(20)=(20-22.5)/1.58~~-1.58` and `z_(24)=(24-22.5)/1.58~~0.95`

-- the `z` score is the number of standard deviations that a given datum is above (positive) or below (negative) the mean; thus a `z` score of -1.58 indicates that 20 is about 1.58 standard deviations below the mean of 22.5

So `P(20<X<24)=P(-1.58<z<0.95)`

Consulting the standard normal table, we find that the area under the curve to the left of `z=-1.58` is .0571 while the area under the curve to the left of 0.95 is .8289. Thus the probability that a z-score is between these values is .8289-.0571=.7718 (a graphing calculator gives .7719)

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**The probability that a pair of batteries lasts between 20 and 24 hours is approximately .7719**

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(2) Find the probability that a randomly selected sample of 20 pairs has a mean life between 21 and 22 hours.

Note that this is a fundamentally different question. For individuals there is great variability; while for samples (at least random samples of sufficient size) there is much less variability. However, we approach the problem in a similar manner.

To convert to a standard score, we use `z=(bar(x)-mu)/(sigma/sqrt(n))` . Note the correction factor in the denominator.

So `z_21=(21-22.5)/(1.58/sqrt(20))~~-4.25` and `z_22=(22-22.5)/(1.58/sqrt(20))~~-1.42`

So `P(21<bar(x)<22)=P(-4.24<z<-1.42)`

Consulting the standard normal table, we see that for `z` below -3.5 we use .0001, while the area beneath the curve to the left of -1.42 is .0778. Thus the area between them is .0778-.0001=.0777 (A graphing calculator gives me .0778 using the approximations for `z` )

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**The probability that the mean of 20 randomly selected samples has a mean between 21 and 22 is .0777**

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