QUESTION 6.Suppose that the mean time for an oil change at \'10-minute oil chang
ID: 3310950 • Letter: Q
Question
QUESTION 6.Suppose that the mean time for an oil change at '10-minute oil change joint' is 11.4 minutes with a standard deviation of 3.2 minutes.
If a random sample of 25 oil changes is selected, what is the probability the mean time will be great than 12 minutes?
0.8264
0.1736
0.7275
Not enough information to solve.
QUESTION 7.Suppose that the mean time for an oil change at '10-minute oil change joint' is 11.4 minutes with a standard deviation of 3.2 minutes.
If a random sample of 36 oil changes is selected, what is the probability the mean time will be great than 12 minutes?
Hint: you'll need to use the Standard Normal Table and Z-Score here. Also, notice the sample size changes to 36! Depending on how you round the standard error your result might be slightly different than the values below. Select the value closest to your solution.
1.13
0.8708
0.1292
Not enough information to answer.
QUESTION 8.Suppose that the mean time for an oil change at '10-minute oil change joint' is 11.4 minutes with a standard deviation of 3.2 minutes.
If a random sample of 25 oil changes is selected, what is the probability the mean time will be great than 10 minutes?
0.0143
(0.9857
0.7500
-2.19
(a)0.8264
(b)0.1736
(c)0.7275
(d)Not enough information to solve.
Explanation / Answer
QUESTION 6, mean is 11.4 and s is 3.2
for sample size N=25, the standard error is SE=s/sqrt(N)=3.2/sqrt(25)=0.64
z=(xbar-mean)/SE
thus P(xbar>12)=P(z>(12-11.4)/0.64)=P(z>0.94) or 1-P(z<0.94)
from normal distribution table we get 1-0.8264=0.1736 option B
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