Question

17#The producer of a weight-loss pill advertises that people who use the pill lose, after one week, an average (mean) of 1.75 pounds with a standard deviation of 0.95 pounds. In a recent study, a group of 60 people who used this pill were interviewed. The study revealed that these people lost a mean of 1.73 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 60 individuals will be 1.73 pounds or less?Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.AGAIN Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

Answer 1

Mean (μ) = 1.75 pounds

Standard deviation (σ) = 0.95 pounds

Sample size (n) = 60

First, we define the z-score for the sample mean distribution:

`Z=\frac{\bar{X}-\mu}{\sigma/√(n)}`

If the mean of a sample of 60 people is 1.73 pounds, the corresponding z-score is:

`Z=(1.73-1.75)/(0.95/√(60))=-0.163073`

Then, the probability that the mean weight loss after one week on the pill for a random sample of 60 individuals will be 1.73 pounds or less is equivalent to:

`P(\bar{X}\le1.73\text{ pounds})=P(Z\le-0.163073)`

Finally, using the standard normal distribution:

`P(Z\leqslant-0.163073)=0.4352`