statistics quick start company mean life and standard deviation

Question

Quick Start Company makes 12-volt car batteries. From historical data, the company knows that the life of such a battery is a normally distributed random variable with a mean life of 40 months and a standard deviation of 8 months.

(a) What percentage of Quick Start 12-volt batteries will last between 33 months and 52 months?

---

a) I used the formula z = x - mean / standard deviation

(33 - 40) / 8 < z < (52 - 40) / 8

-0.875 < z < 1.5

z = -2.375

Usage of z table I got 0.0089

= 0.0089 * 100%

= 0.89%

---

Quick Start Company makes 12-volt car batteries. From historical data, the company knows that the life of such a battery is a normally distributed random variable with a mean life of 40 months and a standard deviation of 8 months.

(d) Seventy-five 12-volt batteries are randomly selected n=75. What is the probability that the mean lifetime of the batteries in this sample will be between 38 and 39 months?

I have used the following formula, not sure if it's correct

z = sample mean - mean / (standard deviation / sqrt n)

z = (38 - 40 / (8 / sqrt 75)) < z < (39 - 40 / (8 / sqrt 75))

z = (-0.028867513) - (-0.014433757)

z = -0.014433756

Probability equals 0.4960 from the z table

I have gotten both of these wrong what are the correct answers?

Answer 1

Checking (a) because I can't follow your logic and the answer seems too small. You have $X \sim \mathsf{Norm}(\mu=40,\,\sigma=8)$ and you seek $P(33 < X \le 52).$ In R statistical software I get the answer $0.7427,$ in agreement with @K.Brix's simulation just posted. You should be able to get close to this same result using standardization and printed tables of the standard normal CDF. So please look at that again.

diff(pnorm(c(33,52), 40, 8))
## 0.0.7424058

Here is a plot of the PDF of $\mathsf{Norm}(\mu=40,\,\sigma=8).$ The probability you want corresponds to the area between the vertical red lines. You should learn to make rough versions of such plots to avoid answers that are very far from correct.

In part (d), you have $\bar X \sim \mathsf{Norm}(40, 8/\sqrt{75})$ and you seek $P(38 < \bar X \le 39).$ From R, I get $0.1243.$ A relevant plot is shown below.

The horizontal scale is the same as in the previous plot, and the thin black curve is again the PDF of the population distribution $\mathsf{Norm}(40, 8).$ The density function of $\bar X$ is the heavy blue curve. It is roughly $\frac{1}{9}$th as 'wide', so it needs to be roughly $9$ times as 'tall' in order for the area under this curve to be unity.

One difficulty with your computation is that (39 - 40 / (8 / sqrt 75)) needs to be $\frac{39 - 40}{8/\sqrt{75}} = -1.0825.$ Another is that you are writing $z =$ a whole bunch of things, only some of which are correct.

I suggest the following format:

$$P(38 < \bar X \leq 39) = P\left(\frac{38-40}{8/\sqrt{75}} < \frac{\bar X - \mu_{\bar X}}{\sigma_{\bar X}} < \frac{39-40}{8/\sqrt{75}}\right) = \cdots = ??$$ where $Z = \frac{\bar X - \mu_{\bar X}}{\sigma_{\bar X}} \sim \mathsf{Norm}(0,1).$

Note: Ordinarily, I might finish this computation for you, but I'm concerned about your overall progress in your statistics course, and it seems to me you are outsourcing way too much of your homework with way too little careful thought on your own part. Please try finishing parts (a) and (d) correctly, and let us know your progress.

Finally, you asked @K.Brix the method of his/her answers. I think the attempt was to give you hints so you would know when you get the right answers, using the methods of your text or lectures. The simulation for (d) might have been something like the following one in R statistical software:

a = replicate(10^6, mean(rnorm(75, 40, 8))) # 1,000,000 averages of samples of size 75
mean(a > 38 & a < 39)
## 0.124804  # Agrees with exact computation to three places.