Question

Harry Potter books have become popular with children and adults alike. A recent survey conducted in London revealed that 64% high school students have read the first Harry Potter book. A random sample of 12 London high school students is taken, and the number of students who have read the first Harry Potter book is recorded
a) Define the random variable of interest and give its distribution, including the values of all the parameters x- sample size of London high school students who participated in the survey xBinomial (iF 12·여 ·64 ) (Round all probability answers to four decimals)
b) What is the expected number of randomly selected students who have read the first Harry Potter book?
c) What is the variance of number of students who have read the first Harry Potter book?
d) What is the probability that exactly two of the randomly selected students has read the first Harry Potter book?
e) What is the probability that at least two of the randomly selected students have read the first Harry Potter book?
f) What is the probability that no more than five of the randomly selected students have read the first Harry Potter book?
g) What is the probability that between two and seven (inclusive) of the randomly selected students have read the first Harry Potter book?
h) What is the probability that more than seven of the randomly selected students have read the first Harry Potter book?

Answer 1

Answer:

a) Let X the random variable of interest "Number of students who have read the first Harry Potter book", on this case we now that:

$X \sim Binom(n=12, p=0.64)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

b) $E(X) = \mu_X = np = 12*0.64= 7.68$

c) $Var(X) = \sigma_X = np(1-p) = 12*0.64(1-0.64)= 2.76$

d) $P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988$

e) $P(X\geq 2) =1- [0.00000474+0.000101]=0.9999$

f)$P(X>5) = 1- P(X \leq 5)=0.9030$

And since we want "NO more than 5" we can do this 1-0.9030= 0.097

g) $P(2\leq X\leq 7) =0.4459$

h) $P(X>7) = 1-P(X\leq 7)=1-0.4458=0.5541$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Part a

Let X the random variable of interest "Number of students who have read the first Harry Potter book", on this case we now that:

$X \sim Binom(n=12, p=0.64)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part b

The expected value is given by the following expression:

$E(X) = \mu_X = np = 12*0.64= 7.68$

Part c

The variance is given by the following expression:

$Var(X) = \sigma_X = np(1-p) = 12*0.64(1-0.64)= 2.76$

Part d

For this case we want this probability

$P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988$

Part e

For this case we want this probability:

$P(X\geq 2) = 1-P(X$

$P(X=0)=(12C0)(0.64)^0 (1-0.64)^{12-0}=0.00000474$

$P(X=1)=(12C1)(0.64)^1 (1-0.64)^{12-1}=0.000101$

$P(X\geq 2) =1- [0.00000474+0.000101]=0.9999$

Part f

For this case we can find this probability first:

$P(X>5) = 1- P(X \leq 5)$

$P(X=0)=(12C0)(0.64)^0 (1-0.64)^{12-0}=0.00000474$

$P(X=1)=(12C1)(0.64)^1 (1-0.64)^{12-1}=0.000101$

$P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988$

$P(X=3)=(12C3)(0.64)^3 (1-0.64)^{12-3}=0.0059$

$P(X=4)=(12C4)(0.64)^4 (1-0.64)^{12-4}=0.0234$

$P(X=5)=(12C5)(0.64)^5 (1-0.64)^{12-5}=0.0666$

$P(X>5) = 1- P(X \leq 5)=0.9030$

And since we want "NO more than 5" we can do this 1-0.9030= 0.097

Part g

For this case we can find this probability first:

$P(2\leq X\leq 7)$

$P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988$

$P(X=3)=(12C3)(0.64)^3 (1-0.64)^{12-3}=0.0059$

$P(X=4)=(12C4)(0.64)^4 (1-0.64)^{12-4}=0.0234$

$P(X=5)=(12C5)(0.64)^5 (1-0.64)^{12-5}=0.0666$

$P(X=6)=(12C6)(0.64)^6 (1-0.64)^{12-6}=0.1382$

$P(X=7)=(12C7)(0.64)^7 (1-0.64)^{12-7}=0.2106$

Adding all the values we got:

$P(2\leq X\leq 7) =0.4457$

Part h

For this case we can find this probability first:

$P(X>7) = 1-P(X\leq 7)$

$P(X=0)=(12C0)(0.64)^0 (1-0.64)^{12-0}=0.00000474$

$P(X=1)=(12C1)(0.64)^1 (1-0.64)^{12-1}=0.000101$

$P(X=2)=(12C2)(0.64)^2 (1-0.64)^{12-2}=0.000988$

$P(X=3)=(12C3)(0.64)^3 (1-0.64)^{12-3}=0.0059$

$P(X=4)=(12C4)(0.64)^4 (1-0.64)^{12-4}=0.0234$

$P(X=5)=(12C5)(0.64)^5 (1-0.64)^{12-5}=0.0666$

$P(X=6)=(12C6)(0.64)^6 (1-0.64)^{12-6}=0.1382$

$P(X=7)=(12C7)(0.64)^7 (1-0.64)^{12-7}=0.2106$

$P(X>7) = 1-P(X\leq 7)=1-0.4458=0.5541$