Question

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, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below? this will be the goal established by the manager.

Answer 1

Answer:

Incomplete question

First aspect of the question is typed below.

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 2l.2-minutes, and the standard deviation 3.5 minutes.

Explanation:

Employees bonus $50

35 oil changes between 10:00 am to 12:00pm

n = 35

10% changes

So, the z - score can be calculated using

z-score = InvNorm(0.10)

z-score = -1.28

So, given that,

Standard deviation is 3.5minutes

σ = 3.5 minutes

Mean time is 21.2 minutes

μx = 21.2 minutes

Then,

σx = σ / √n

σx = 3.5 / √35

σx = 0.5916 minutes

Then, Z score can be written as

Z = (x - μx) / σx

-1.28 =  (x - 21.2) / 0.5916

Cross multiply

-1.28 × 0.5916 = x - 21.2

-0.7573 = x - 21.2

x = 21.2 - 0.7573

x = 20.443 minutes

There is a 10% chance of being at or below a mean oil-change time of 20.44 minutes