Question

The producer of a weight-loss pill advertises that people who use the pill lose, after one week, an average (mean) of pounds with a standard deviation of pounds. In a recent study, a group of people who used this pill were interviewed. The study revealed that these people lost a mean of pounds after one week. If the producer's claim is correct, what is the probability that the mean weight loss after one week on this pill for a random sample of individuals will be pounds or less?

Answer 1

Answer:

$P(\bar X <1.92)=P(Z<0.847)=0.802$

Step-by-step explanation:

Assuming this info: The producer of a weight-loss pill advertises that people who use the pill lose, after one week, an average (mean) 1.8 of pounds with a standard deviation of 0.95 pounds. In a recent study, a group of 45 people who used this pill were interviewed. The study revealed that these people lost a mean of 1.92 pounds after one week. If the producer's claim is correct, what is the probability that the mean weight loss after one week on this pill for a random sample of individuals will be 1.92 pounds or less?

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the interpupillary distance (the distance between the pupils of the left and right eyes) of a population, and for this case we know the distribution for X is given by:

$X \sim N(1.8,0.95)$  

Where $\mu=1.8$ and $\sigma=0.95$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case  $\bar X \sim N(1.8,\frac{0.95}{\sqrt{45}})$

We are interested on this probability

$P(\bar X\leq 1.92)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

2) Solution to the problem

If we apply this formula to our probability we got this:

$P(\bar X\leq 1.92)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}\leq \frac{1.92-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(Z<\frac{1.92-1.8}{\frac{0.95}{\sqrt{45}}})=P(Z<0.847)$

And we can find this probability on this way:

$P(Z<0.847)=0.802$

We can use the following excel code to find it:

"=NORM.DIST(0.847,0,1,TRUE)"