Question

shape of distribution of time required to get oil change is unknown. records indicate mean time 11.5 and standard deviation 3.9. To compute probabilities regarding the sample mean using normal model what size sample would be required? suppose manager agrees pay each employee bonus if they meet goal. typical Saturday facility perform 35 oil changes. treating this as random sample what mean would there be 10% chance of being at or below?

Answer 1

shape of distribution of time required to get oil change is unknown. records indicate mean time 11.5 and standard deviation 3.9.

=11.5    = 3.9

  

To compute probabilities regarding the sample mean using normal model what size sample would be required?

Usually when the sample increases that is getting larger, shape of distribution of mean takes a shape of normal distribution. Large sample means 30 or more.

Therefore, to use the normal model we require a sample size of at least 30.

suppose manager agrees pay each employee bonus if they meet goal. typical Saturday facility perform 35 oil changes. treating this as random sample what mean would there be 10% chance of being at or below?

n =35 =11.5    = 3.9

Since n > 30, we can use the normal model for the distribution of mean.

Here we have to find the probability of the average so we will need the distribution of the mean. We will use the central limit theorem which states that if population is normal then the distribution of the means of its samples is also normal. The parameters are as follows.

(11.5 , 0.65922)

z-score = ( - 11.5) / 0.65922

Here we want a value where the sample mean is below or at with a probability of 10%.

P( < a) = 10%

So we need to find the z-score from probability.  We use normal percentage tables for this. The tables provide value for greater than for 'p < 50%' . We would have to rearrange.

(P( > a) = 90%

P( > - a) = 10%

z-score = 1.28155

-a = 1.28155

( - 11.5) / 0.65922 = -1.28155

= 10.6552