The rest of question C continued ... : there is a 10% chance at being at or belo

Question

The rest of question C continued ... : there is a 10% chance at being at or below a mean oil change time of _____ minutes . The shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is unknown However, records indicate that the mean time is 21.7 minutes, and the standard deviation is 4.8 minutes. Complete parts (a) through (c). (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? OA. The normal model cannot be used if the shape of the distribution is unknown B. OC. OD. The sample size needs to be less than or equal to 30. The sample size needs to be greater than or equal to 30. Any sample size could be used. (b) What is the probability that a random sample of n 35 oil changes results in a sample mean time less than 20 minutes? The probability is approximately (Round to four decimal places as needed.) (e) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday the oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? This will be the goal Click to select your answer(s)

Explanation / Answer

a)

The sample needs to be large enough, n >= 30. Hence,

OPTION C. [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    20      
u = mean =    21.7      
n = sample size =    35      
s = standard deviation =    4.8      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.095278257      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.095278257   ) =    0.018073132 [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.10      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    21.7      
z = the critical z score =    -1.281551566      
s = standard deviation =    4.8      
n = sample size =    35      
Then          
          
x = critical value =    20.66021559 minutes [ANSWER]      

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