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 related to the Z score.
To answer this question, we will use the concept of sampling distribution and the central limit theorem.
The central limit theorem states that for a large enough sample size, the distribution of sample means will be approximately normally distributed regardless of the shape of the population distribution.
In this case, we are given the population mean (µ = 175.5) and the population standard deviation (σ = 90.57). We want to find the probability that the mean number of Facebook friends in a sample of 25 students is 190 friends or more (x̄ ≥ 190).
First, we need to calculate the standard deviation of the sampling distribution, also known as the standard error (SE). The formula for the standard error is:
SE = σ / √n
Where σ is the population standard deviation and n is the sample size.
In our case, σ = 90.57 and n = 25, so:
SE = 90.57 / √25 = 90.57 / 5 = 18.114
Next, we convert the sample mean (190) to a z-score using the formula:
z = (x - µ) / SE
Where x is the sample mean and µ is the population mean.
z = (190 - 175.5) / 18.114 ≈ 0.80
Now, we need to use a standard normal distribution table or a calculator to find the probability associated with this z-score.
Looking up the z-score of 0.80 in a standard normal distribution table, we find that the probability is approximately 0.789.
Therefore, the probability that the mean number of Facebook friends in a sample of 25 students will be 190 friends or more is approximately 0.789 or 78.9%.