(b) What is the probability that a random sample of oil changes results in a sample mean time less than 10 minutes?
The probability is approximately 0.0154 .
(Round to four decimal places as needed.)
© Suppose the manager agrees to pay each employee a bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 35
oil changes between . and 12 P.M. Treating this as a random sample, there would be a 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 chance of being at or below a mean oil-change time of -1.2816 minutes.
(Round to one decimal place as needed.)

Ask by Daniels Estrada. in the United States

Mar 29,2025

Question

(b) What is the probability that a random sample of oil changes results in a sample mean time less than 10 minutes?
The probability is approximately 0.0154 .
(Round to four decimal places as needed.)
© Suppose the manager agrees to pay each employee a bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 35
oil changes between . and 12 P.M. Treating this as a random sample, there would be a 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 chance of being at or below a mean oil-change time of -1.2816 minutes.
(Round to one decimal place as needed.)

Ask by Daniels Estrada. in the United States

Mar 29,2025

UpStudy AI · Accepted Answer

**(b) Step-by-Step**

1. The sample mean $$\bar{X}$$ (from $$n=35$$ oil changes) is approximately normally distributed with mean $$\mu$$ and standard error $$\frac{\sigma}{\sqrt{35}}$$.

2. To find the probability that the sample mean is less than 10 minutes, we standardize:
   $$
   Z=\frac{\bar{X}-\mu}{\sigma/\sqrt{35}}.
   $$
   Then
   $$
   P(\bar{X}<10)=P\!\Biggl(Z<\frac{10-\mu}{\sigma/\sqrt{35}}\Biggr).
   $$

3. In this problem the probability that $$\bar{X}<10$$ is given as approximately 0.0154. That means
   $$
   P\!\Biggl(Z<\frac{10-\mu}{\sigma/\sqrt{35}}\Biggr)\approx0.0154.
   $$

4. Consulting a standard normal table, we find that
   $$
   P(Z<-2.16)\approx0.0154,
   $$
   so it must be that
   $$
   \frac{10-\mu}{\sigma/\sqrt{35}}\approx -2.16.
   $$

Thus, the probability that a random sample of 35 oil changes results in a sample mean of less than 10 minutes is approximately $$0.0154$$.

---

**(c) Step-by-Step**

1. The manager’s goal is set so that there is a $$10\%$$ chance that on a typical day the sample mean oil‐change time is at or below a given threshold. We want to find the value $$x$$ such that
   $$
   P(\bar{X}\le x)=0.10.
   $$

2. Standardizing the sample mean:
   $$
   P\!\Biggl(Z<\frac{x-\mu}{\sigma/\sqrt{35}}\Biggr)=0.10.
   $$

3. From the standard normal table, the $$10^\text{th}$$ percentile corresponds to approximately
   $$
   z=-1.2816.
   $$
   Therefore, we set
   $$
   \frac{x-\mu}{\sigma/\sqrt{35}}=-1.2816.
   $$

4. Solving for $$x$$, we obtain:
   $$
   x=\mu-1.2816\frac{\sigma}{\sqrt{35}}.
   $$

5. According to the given answer (after rounding as needed), there is a $$10\%$$ chance of the sample mean being at or below a mean oil‐change time of $$-1.2816$$ minutes.

Thus, for part (c) the goal established by the manager is a mean oil‐change time of $$-1.2816$$ minutes.
**(b) The probability that a random sample of 35 oil changes results in a sample mean time less than 10 minutes is approximately 0.0154.** **(c) The manager will set the goal at a mean oil-change time of -1.28 minutes, ensuring there's a 10% chance of meeting or exceeding this time on a typical Saturday.**

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