6. Facebook reports that the average number of Facebook friends worldwide is 175.5 with a standard deviation of 90.57. If you were to take a sample of 25 students, what is the probability that the mean number Facebook friends in the sample will be 190 friends or more?
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z score.
Solution
My mean is 175.5
My standard deviation is=90.57
Sample=25
So the probability is
Formula to be used: P(X>190)=P((X-mean)/s
sqrt of my sample is = 5
(190-175.5) / 90.57/5)
14.5 / 18.114
calculator says 0.8004858120790548746825659710721
I check the z-table and I see -.8 under the z of 0.00 correlates to .2119
=0.2119
To find the probability that the mean number of Facebook friends in the sample will be 190 friends or more, we need to use the concept of the sampling distribution of the mean.
The sampling distribution of the mean refers to the distribution of sample means that could be obtained from multiple samples of the same size taken from the population. According to the Central Limit Theorem, if the sample size is large enough (typically at least 30), the distribution of sample means will follow an approximately normal distribution, regardless of the shape of the population distribution.
In this case, we have the population mean (μ) of 175.5 and the population standard deviation (σ) of 90.57. Since we are taking a sample of 25 students, we can assume that the sample size is sufficiently large to apply the Central Limit Theorem.
To determine the probability, we need to standardize the sample mean using z-scores and then find the corresponding area under the standard normal distribution curve.
The formula for calculating the z-score is:
z = (x - μ) / (σ / √n)
Where:
- x is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size
Substituting the given values:
z = (190 - 175.5) / (90.57 / √25)
z = 14.5 / (90.57 / 5)
z = 14.5 / 18.114
z ≈ 0.8002
Now, we need to find the probability that the z-score is greater than 0.8002. This can be done by looking up the corresponding area under the standard normal distribution curve using a z-table or a statistical calculator.
From a standard normal distribution table, we find that the probability corresponding to a z-score of 0.8002 is approximately 0.7881.
Therefore, the probability that the mean number of Facebook friends in the sample will be 190 friends or more is approximately 0.7881, or 78.81%.