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?
To solve this problem, we can use the Central Limit Theorem and the properties of the normal distribution.
The Central Limit Theorem states that for a large enough sample size (greater than 30), the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution. In this case, we are given a sample size of 25, which is considered large enough to apply the Central Limit Theorem.
First, let's calculate the sampling distribution's mean (also known as the expected value) and standard deviation using the given information. The mean of the sampling distribution is equal to the mean of the population, which is 175.5. The standard deviation of the sampling distribution is calculated by dividing the population standard deviation by the square root of the sample size:
Standard deviation of the sampling distribution (σ) = population standard deviation / √sample size
= 90.57 / √25
= 90.57 / 5
= 18.114
Now, let's convert the probability question into a Z-score, which represents the number of standard deviations away from the mean.
Z = (sample mean - population mean) / (standard deviation / √sample size)
Z = (190 - 175.5) / (18.114)
Z = 14.5 / 18.114
Z ≈ 0.8004
The next step is to find the probability of Z being greater than 0.8004. We can use a standard normal distribution table or a calculator to find this probability. The table gives us the area under the curve to the left of a given Z-score, so we need to subtract the obtained probability from 1 to get the area to the right.
Using a standard normal distribution table or a calculator, the probability of Z > 0.8004 is approximately 0.2119.
Therefore, the probability that the mean number of Facebook friends in the sample will be 190 friends or more is approximately 0.2119, or 21.19%.