If the mean number of cigarettes smoked by pregnant women is 12 and the standard deviation is 8, find the probability that in a random sample of 100 pregnant women the mean number of cigarettes smoked will be greater than 20. What does this mean (within your population).
To find the probability that the mean number of cigarettes smoked by a random sample of 100 pregnant women will be greater than 20, we can use the central limit theorem along with the given information to approximate the sample mean distribution as a normal distribution.
The mean number of cigarettes smoked by pregnant women is given as 12, and the standard deviation is 8. Since our sample size is large (100), we can assume that the sample mean will follow a normal distribution.
To find the probability, we need to standardize the sample mean and calculate the area under the standard normal curve.
First, we need to find the standard error of the mean (SE):
SE = standard deviation / square root of sample size
SE = 8 / sqrt(100)
SE = 8 / 10
SE = 0.8
Now, let's standardize the value of 20:
Z = (sample mean - population mean) / SE
Z = (20 - 12) / 0.8
Z = 10 / 0.8
Z = 12.5
Using a standard normal distribution table or a calculator, we can find the probability associated with a Z-score of 12.5. However, since the Z-score is quite large, the probability will be extremely small.
So, the probability that in a random sample of 100 pregnant women, the mean number of cigarettes smoked will be greater than 20 is approximately 0.
This means that within the population of pregnant women, it is highly unlikely to find a random sample of 100 women where the mean number of cigarettes smoked is greater than 20. Most pregnant women in the population smoke fewer cigarettes than that.