Question

marcus reads in a magazine article that 62% of teenagers prefer batman over iron man. he decides to test this so he takes a simple random sample of 50 people at wal - mart on a saturday afternoon.

1. what is the probability that he finds exactly 40 batman fans?

2nd, distr, binomial ()=

1. what is the probability that he finds fewer than 30 batman fans?

2nd, distr, binomial ()=

1. what is the probability that marcus will find more than 35 batman fans?

2nd, distr, binomial ()=
suppose that 20% of the otters in a large aquatic community are typically infected with a parasite. a biologist samples 16 otters at random and records whether or not the animal is infected.

1. find the expected number of infected otters in the sample.
2. what is the probability exactly 4 of the sampled otters are infected?

2nd, distr, binomial ()=

1. what is the probability that 8 or more of the sampled otters are infected?

1 - 2nd, distr, binomial ()=

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: For problem 15

Here, \(n = 50\), \(k = 40\), \(p=0.62\), \(1-p = 0.38\)
\(C(50,40)=\frac{50!}{40!(50 - 40)!}=\frac{50!}{40!×10!}=\frac{50\times49\times\cdots\times41}{10\times9\times\cdots\times1}\)
\(P(X = 40)=C(50,40)\times(0.62)^{40}\times(0.38)^{10}\)
Using a binomial probability calculator (as \(2^{nd}\text{, distr, binomial}(n = 50,p = 0.62,k = 40)\))

Step3: For problem 16

\(P(X\lt30)=\sum_{k = 0}^{29}C(50,k)\times(0.62)^{k}\times(0.38)^{50 - k}\)
Using a binomial cumulative - distribution calculator (\(2^{nd}\text{, distr, binomial}(n = 50,p = 0.62,k = 29)\))

Step4: For problem 17

\(P(X\gt35)=1 - P(X\leq35)\)
\(P(X\leq35)=\sum_{k = 0}^{35}C(50,k)\times(0.62)^{k}\times(0.38)^{50 - k}\)
Using a binomial cumulative - distribution calculator and then subtracting from 1 (\(1-2^{nd}\text{, distr, binomial}(n = 50,p = 0.62,k = 35)\))

Step5: For problem 18

The expected value of a binomial distribution \(E(X)=np\)
Here, \(n = 16\), \(p = 0.2\)
\(E(X)=16\times0.2 = 3.2\)

Step6: For problem 19

\(n = 16\), \(k = 4\), \(p = 0.2\), \(1 - p=0.8\)
\(C(16,4)=\frac{16!}{4!(16 - 4)!}=\frac{16\times15\times14\times13}{4\times3\times2\times1}=1820\)
\(P(X = 4)=C(16,4)\times(0.2)^{4}\times(0.8)^{12}\)
Using a binomial probability calculator (\(2^{nd}\text{, distr, binomial}(n = 16,p = 0.2,k = 4)\))

Step7: For problem 20

\(P(X\geq8)=1 - P(X\leq7)\)
\(P(X\leq7)=\sum_{k = 0}^{7}C(16,k)\times(0.2)^{k}\times(0.8)^{16 - k}\)
Using a binomial cumulative - distribution calculator and then subtracting from 1 (\(1 - 2^{nd}\text{, distr, binomial}(n = 16,p = 0.2,k = 7)\))

Answer:

1. Using a binomial probability calculator: \(P(X = 40)\approx0.0003\)
2. Using a binomial cumulative - distribution calculator: \(P(X\lt30)\approx0.113\)
3. Using a binomial cumulative - distribution calculator: \(P(X\gt35)\approx0.137\)
4. \(E(X)=3.2\)
5. Using a binomial probability calculator: \(P(X = 4)\approx0.200\)
6. Using a binomial cumulative - distribution calculator: \(P(X\geq8)\approx0.005\)