# What is the probability that one random selected pair of batteries will have a test life between 20 and 24 hours?  Standard normal distribution question: A popular flashlight that uses two D-size batteries was selected, and several of the same models were purchased to test the "continous use-life" of D batteries. To determine the flashlight, the flashlight was turned on and the time noted. When the flashlight no longer produced light, the time was again noted. The resulting "life" data from the batteries had a mean of 22.5 hours. Assume these values have a normal distribution with a standard deviation of 1.58 hours. A. What is the probability that one random selected pair of batteries will have a test life between 20 and 24 hours? B. What is the probability that a randomly selected sample of 20 pairs of batteries will have a mean test life between 21 and 22 hours?

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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