Question

suppose that the average number of facebook friends users have is normally distributed with a mean of 117 and a standard deviation of about 55. assume forty - one individuals are randomly chosen. answer the following questions. round all answers to 4 decimal places where possible.
a. what is the distribution of \\(\bar{x}\\)? \\(\bar{x}\sim n(117,8.5896)\\)
b. for the group of 41, find the probability that the average number of friends is more than 130.
0.0660
c. find the first quartile for the average number of facebook friends.

d. for part b), is the assumption that the distribution is normal necessary? no yes
hint:
some helpful videos:

• finding the sampling distribution +
• finding a probability using the central limit theorem +
• finding value given a probability using the central limit theorem +
• the central limit theorem for sums +

hint

Explanation:

Step1: Recall the formula for the standard deviation of the sample - mean

The standard deviation of the sample - mean (also known as the standard error) is given by $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Given $\sigma = 55$ and $n = 41$, then $\sigma_{\bar{x}}=\frac{55}{\sqrt{41}}\approx8.5896$. The distribution of $\bar{X}$ is $\bar{X}\sim N(\mu,\sigma_{\bar{x}})$, with $\mu = 117$ and $\sigma_{\bar{x}}\approx8.5896$.

Step2: Standardize the value for part b

To find $P(\bar{X}>130)$, we first standardize the value using the formula $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x} = 130$, $\mu = 117$, and $\sigma_{\bar{x}}\approx8.5896$. So $z=\frac{130 - 117}{8.5896}=\frac{13}{8.5896}\approx1.5135$. Then $P(\bar{X}>130)=P(Z > 1.5135)=1 - P(Z\leq1.5135)$. From the standard - normal table, $P(Z\leq1.5135)\approx0.9345$, so $P(\bar{X}>130)=1 - 0.9345 = 0.0655$.

Step3: Find the z - score for the first quartile

The first quartile corresponds to a cumulative probability of $P(Z)=0.25$. Looking up in the standard - normal table, the $z$ - score $z_{0.25}\approx - 0.6745$.

Step4: Use the z - score formula to find the first quartile for $\bar{X}$

We know that $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, so $\bar{x}=\mu+z\sigma_{\bar{x}}$. Substituting $\mu = 117$, $z=-0.6745$, and $\sigma_{\bar{x}}\approx8.5896$, we get $\bar{x}=117+( - 0.6745)\times8.5896=117-5.7949\approx111.2051$.

Step5: Determine the necessity of the normal assumption for part b

Since $n = 41\geq30$, by the Central Limit Theorem, the sampling distribution of the sample mean $\bar{X}$ is approximately normal regardless of the shape of the population distribution. So the answer is No.

Answer:

a. $\bar{X}\sim N(117,8.5896)$
b. $0.0655$
c. $111.2051$
d. No