(c) 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 40 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 established by the manager. There is a 10% chance of being at or below a mean oil-change time of minutes.

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How would I find the answer with my TI-84?
I have no idea what to do after getting P(Z< -1.55).....?


Mean is 11.1
Standard Deviation: 4.5
n = 40

I got so far: 10-11.1 = -1.1 / 4.5 divided by the square root of 40

so around : -1.55

(c) 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 40 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 established by the manager.
There is a 10% chance of being at or below a mean oil-change time of
(Round to one decimal place as needed.)
minutes.
Transcribed Image Text:(c) 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 40 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 established by the manager. There is a 10% chance of being at or below a mean oil-change time of (Round to one decimal place as needed.) minutes.
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