Question

1. ten percent of college graduates think that judge judy serves on the supreme court. you randomly select five college graduates and ask them whether they think that judge judy serves on the supreme court. the random variable represents the number of college graduates who think that judge judy serves on the supreme court.

a. find the mean, variance, and standard deviation of the binomial distribution for the given random variable. interpret the results.

b. find the probability that exactly 2 of the five randomly selected college graduates think judge judy serves on the supreme court.

c. find the probability that at least 2 of the five randomly selected college graduates think judge judy serves on the supreme court.

d. find the probability that at most 2 of the five randomly selected college graduates think judge judy serves on the supreme court.

Explanation:

Step1: Identify parameters for binomial distribution

For a binomial distribution \(X\sim B(n,p)\), where \(n = 5\) (number of trials, i.e., number of college graduates selected) and \(p=0.1\) (probability of success, i.e., probability that a college graduate thinks Judge Judy serves on the Supreme Court).

Step2: Calculate the mean

The formula for the mean of a binomial distribution is \(\mu=np\).
Substituting \(n = 5\) and \(p = 0.1\), we get \(\mu=5\times0.1 = 0.5\).

Step3: Calculate the variance

The formula for the variance of a binomial distribution is \(\sigma^{2}=np(1 - p)\).
Substituting \(n = 5\), \(p = 0.1\), and \(1-p=0.9\), we get \(\sigma^{2}=5\times0.1\times0.9=0.45\).

Step4: Calculate the standard deviation

The formula for the standard deviation is \(\sigma=\sqrt{np(1 - p)}\).
Since \(\sigma^{2}=0.45\), \(\sigma=\sqrt{0.45}\approx0.67\).

Step5: Calculate the probability for part b

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).
For \(n = 5\), \(k = 2\), \(p = 0.1\), and \(1-p=0.9\):
\(C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4}{2\times1}=10\)
\(P(X = 2)=10\times(0.1)^{2}\times(0.9)^{3}=10\times0.01\times0.729 = 0.0729\)

Step6: Calculate the probability for part c

\(P(X\geq2)=1-(P(X = 0)+P(X = 1))\)
\(P(X = 0)=C(5,0)\times(0.1)^{0}\times(0.9)^{5}=1\times1\times0.59049 = 0.59049\)
\(C(5,1)=\frac{5!}{1!(5 - 1)!}=5\)
\(P(X = 1)=5\times(0.1)^{1}\times(0.9)^{4}=5\times0.1\times0.6561 = 0.32805\)
\(P(X\geq2)=1-(0.59049 + 0.32805)=1 - 0.91854=0.08146\)

Step7: Calculate the probability for part d

\(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\)
We know \(P(X = 0)=0.59049\), \(P(X = 1)=0.32805\), \(P(X = 2)=0.0729\)
\(P(X\leq2)=0.59049+0.32805 + 0.0729=0.99144\)

Answer:

a. Mean \(\mu = 0.5\), Variance \(\sigma^{2}=0.45\), Standard deviation \(\sigma\approx0.67\). On average, 0.5 out of 5 college graduates think Judge Judy serves on the Supreme Court. The spread (variance) around this mean is 0.45 and the standard deviation (a measure of spread) is approximately 0.67.
b. \(P(X = 2)=0.0729\)
c. \(P(X\geq2)=0.08146\)
d. \(P(X\leq2)=0.99144\)