Question

given the following sampling distribution: what is the mean of this sampling distribution? p(x) x 1/50 -17 9/100 -9 1/20 -5 9/100 8 13 o 9.2 o 8.8 o 9.1 o -2.0 o -0.3 o none of the above

Explanation:

Step1: Recall mean formula for discrete - distribution

The mean $\mu$ of a discrete probability distribution is given by $\mu=\sum_{i}x_ip_i$, where $x_i$ are the values of the random - variable and $p_i$ are their corresponding probabilities.

Step2: Calculate the products

For the first row: $x_1=-17$ and $p_1 = \frac{1}{50}=0.02$, so $x_1p_1=-17\times0.02=-0.34$.
For the second row: $x_2=-9$ and $p_2=\frac{9}{100}=0.09$, so $x_2p_2=-9\times0.09 = - 0.81$.
For the third row: $x_3=-5$ and $p_3=\frac{1}{20}=0.05$, so $x_3p_3=-5\times0.05=-0.25$.
For the fourth row: $x_4 = 8$ and $p_4=\frac{9}{100}=0.09$, so $x_4p_4=8\times0.09 = 0.72$.
Let the probability of $x_5 = 13$ be $p_5$. Since the sum of all probabilities in a probability distribution is 1, we have $p_5=1-(0.02 + 0.09+0.05 + 0.09)=1 - 0.25=0.75$. Then $x_5p_5=13\times0.75 = 9.75$.

Step3: Sum the products

$\mu=-0.34-0.81 - 0.25+0.72+9.75$.
First, combine the negative terms: $-0.34-0.81-0.25=-1.4$.
Then, $\mu=-1.4 + 0.72+9.75=-0.68+9.75 = 9.07\approx9.1$.

Answer:

9.1