need help with #5 5. The accumulated miles between repairs for vehicle engines i

Question

need help with #5

5. The accumulated miles between repairs for vehicle engines is 24,000 miles with a standard deviation of 2000 miles. The accumulated miles, which have been recorded over time, follow a normal distribution a. Find the probability that an engine you just received will last longer than 26,000 miles. b. Find the probability that the mean accumulated mileage from a sample of 10 engines exceeds 26,000 miles. c. Find the 1s, 2nd, and 3rd quartiles for the accumulated miles between repairs. d. Now, you are looking at vehicle transmissions. The historical data for transmission mileages indicates a population mean of 16,000 miles with a standard deviation of 2800 miles. The mileage for transmissions does not follow a normal distribution. Find the probability that, in a large train shipment of 40 transmissions, the average mileage for this sample will be less than 15,000 miles. e. If the average for your transmission sample of 40 falls below the bottom 10%, you are going to declare a stand-down of the workforce to determine what is going wrong. What is the cutoff number of miles for the bottom 10% of your sample average. f. Back to the engines. .. If a single engine is considered a "failure" if it doesn't accumulate at least 21,000 miles between repairs, what is the chance that an engine will fail to meet its anticipated mileage accumulation? g. Given the criteria just stated, what would be the "expected number" of failures in the next 1000 engines that are placed into vehicles?

Explanation / Answer

Let X be the random variable that the accumulated miles between repairs for vehicle engines.

Here X have mean (mu) = 24000 miles and

Standard deviation (sd) = 2000 miles

And the population distribution is normal.

z-score for x = 26000 is,

z =(x – mu) / sd = (26000 – 24000) / 2000 = 1

Now we have to find P(Z > 1)

By using statistical table,

P(Z > 1) = 0.1587

2. Given that,

n = 10

Now we have to find P(Xbar > 26000)

z-score for Xbar = 26000 is,

z = (Xbar – mu) / (sd / sqrt(n))

= (26000 – 24000) / (2000 / sqrt(10) = 3.16

Now we have to find P(Z > 3.16)

P(Z > 3.16) = 0.0008

3. Here we have to find 1st , 2nd and 3rd quartile.

In this case we have to find x for 25%, 50% and 75%.

X we can find by using formula,

X = mu + z*sd

Where z is z-score.

Z we can find using statistical table.

First quartile :

Symbolically we can write as,

P(X < x) = 25% = 0.25

For 25% z = -0.67

X = 24000 + (-0.67) * 2000 = 22651.02

Second quartile :

Symbolically we can write as,

P(X < x) = 50% = 0.50

For 50% z = 0.00

X = 24000 + (0.00) * 2000 = 24000

Third quartile :

Symbolically we can write as,

P(X < x) = 75% = 0.75

For 75% z = 0.67

X = 24000 + (0.67) * 2000 = 25348.98

3. In this example we have given,

Mu = 16000 miles

Sd = 2800 miles

n = 40

And we have to find P(Xbar < 15000 miles)

z-score for Xbar = 15000 is,

z = (15000 – 16000) / (2800 / sqrt(40)) = -2.26

Now we have to find P(Z < -2.26)

P(Z < -2.26) = 0.0119

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