Question

question number 2. (10.00 points) given the following sampling distribution: what is p(x > -11)? 0.91 0.94 0.95 0.90 0.92 none of the above

Explanation:

Step1: Recall probability formula

We know that \(P(X > - 11)\) is the sum of probabilities of all \(X\) values greater than \(-11\).

Step2: Identify relevant \(X\) - values

The \(X\) values greater than \(-11\) in the table are \(-3\), \(11\), and \(16\). Their corresponding probabilities are \(\frac{1}{20}\), \(\frac{9}{100}\), and \(1 - (\frac{1}{20}+\frac{3}{100}+\frac{1}{20})\) (since the sum of all probabilities is 1). First, find the sum of the known non - relevant probabilities: \(\frac{1}{20}+\frac{3}{100}=\frac{5 + 3}{100}=\frac{8}{100}\), and \(\frac{8}{100}+\frac{1}{20}=\frac{8+5}{100}=\frac{13}{100}\). So the probability of the last value is \(1-\frac{13}{100}=\frac{87}{100}\).

Step3: Calculate \(P(X > - 11)\)

\(P(X > - 11)=\frac{1}{20}+\frac{9}{100}+\frac{87}{100}\). \(\frac{1}{20}=\frac{5}{100}\), then \(\frac{5}{100}+\frac{9}{100}+\frac{87}{100}=\frac{5 + 9+87}{100}=\frac{101}{100}=1 - \frac{3}{100}= 0.97\). But we made a mistake above. Let's start over. The correct way is \(P(X>-11)=P(X = - 3)+P(X = 11)+P(X = 16)\). First, find \(P(X = 16)=1-(\frac{1}{20}+\frac{3}{100}+\frac{1}{20})=1 - (\frac{5 + 3+5}{100})=1-\frac{13}{100}=\frac{87}{100}\), \(P(X=-3)=\frac{1}{20}=\frac{5}{100}\), \(P(X = 11)=\frac{9}{100}\). \(P(X>-11)=\frac{5}{100}+\frac{9}{100}+\frac{87}{100}=\frac{5 + 9+87}{100}=0.92\).

Answer:

0.92