Question

Once a month, a manufacture selects a sample of solar panels. Each panel is exposed to sunlight for an hour and the amount of electricity emitted is recorded. After several years of sampling, the manufacturer claims that the population mean is 300 watts of power and the population standard deviation is 15 watts of power. An inspection agency decides to test the manufacturer’s claim. The agency takes a sample of 25 panels and tests the electrical output of each panel in the sample

. 2a) Find the probability that the agency’s sample with have a mean electrical output between 294.6 and 305.4 watts of electricity.

2b) Find the probability that the mean of the agency’s sample will be either less than 294.6 watts or greater than 305.4 watts. Hint: Refer to your answer from (2a)

2c) After studying a sample of solar panels, the agency finds that the mean of their sample is 250 watts of power. One of two scenarios occurred: i. The sample contained defective outliers, which caused a sample mean lower than expected. ii. The manufacturer’s claims were false. Referring to your answers to (2a) and (2b), explain which scenario is more likely.

Answer 1

Let $\overline{X}$ be the sample mean of 25 sample panels.

Assuming that the sample mean $\overline{X}$ be normally distributed with mean,

$\overline{X}$~$N(300,\frac{15^{2}}{25})$

2a)

To find

P(294.6<$\overline{X}$<305.4)

$=P(\frac{294.6-300}{15/\sqrt{25}}< \frac{\overline{X}-300}{15/\sqrt{25}} <\frac{305.4-300}{15/\sqrt{25}})$

=P(-1.8<Z<1.8) [where Z~N(0,1) is the standard normal variable]

=0.9281394 [values from the standard normal table]

2b)

to find

P($\overline{X}$<294.6,$\overline{X}$>305.4)

=1-P(294.6<$\overline{X}$<305.4)

=1-0.9281394

=0.0718606

2c)

We have seen that the almost 93% probability lies in P(294.6<$\overline{X}$<305.4), outside is 7% chance.

But the mean is 250 can only be possible that we get a sample mean lower than expected.