1. ## Standard Deviation problem...

At a motor repair shop, a mechanic can tune-up a car in an average of 30 minutes with a standard deviation of 5 minutes. Assume that the time it takes for a tune-up is approximately normally distributed.

(a) If a car arrives for a tune-up 25 minutes before closing, and a mechanic starts work straight away, what is the probability that the car will NOT be tuned-up before closing time?

(b) If four cars arrive 35 minutes before closing, and four mechanics start work straight away, what is the probability that at least one car will not be tuned-up before closing time?

Not really sure where to start so any help would be great! Thanks!

2. Problem (a) involves finding the proportion of the area under the normal curve above 25, given a mean of 30 and s.d. of 5.

Problem (b) is a bit more involved, in that you need to find the area under the curve beyond 35, but once you find that, you need to use the binomial probability distribution to determine the probability that at least 1 car takes longer than 35 minutes (i.e., 1 - prob(no cars take longer than 35 minutes).

If you look in our examples section, you'll find references to the approaches needed to solve these problems.

3. Thanks,

So for a) I would use the z = (x-u)/o) and it would come out as z = (25 -30)/5... Thus the probability would be 0.1587??? Am I doing that right?

4. Also,

Would I approach this question in the same way?

"National Distributing employs 5,000 salespersons throughout Australia. The
average yearly commission earned by a salesperson is \$57,000 and the standard deviation of the yearly commission is \$8,540.

If a random sample of 60 salespersons is selected, what is the probability that the sample mean of their yearly commissions is less than \$55,000?"

So have z = (55,000 - 57,000) / 8540 = -0.234 so the answer would be 0.4090?

5. Originally Posted by Taka
Thanks,

So for a) I would use the z = (x-u)/o) and it would come out as z = (25 -30)/5... Thus the probability would be 0.1587??? Am I doing that right?
Yes, that's right.

6. Originally Posted by Taka
Also,

Would I approach this question in the same way?

"National Distributing employs 5,000 salespersons throughout Australia. The
average yearly commission earned by a salesperson is \$57,000 and the standard deviation of the yearly commission is \$8,540.

If a random sample of 60 salespersons is selected, what is the probability that the sample mean of their yearly commissions is less than \$55,000?"

So have z = (55,000 - 57,000) / 8540 = -0.234 so the answer would be 0.4090?
You have the right idea, but in this particular case, you are determining the distribution of sample means, not individual values, so the denominator should be the s.d. divided by the square root of the sample size:

z = (x-u) / (s/sqrt(n))

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