Question

The probability that an individual is left-handed is 0.12. in a class of 39 students, what is the probability of finding five left-handers

Answer 1

We are given:

p = probability = 0.12
n = total students = 39 

x = left handers = 5
u = mean = p* n = 4.68 
σ = standard dev = √ ( n*p*(1-p)) = √ ( 39 * 0.12 * 0.88 ) = 2.03

 

Calculating for the z score:

z = (x – u) / σ
z = (5 – 4.68) / 2.03

z = 0.1576 = 0.16

Using the standard tables for z, the p value is:

p value = 0.5636 = 56.36%

 

Hence there is a 56.36% chance.

 

Answer 2

Answer:

0.1856

Step-by-step explanation:

Given that prob that an individual is left-handed is 0.12.

n = sample size = no of students = 39

q = 1-p = 0.88

Let X be the no of students who are left handed.

X is binomial since each student is independent of the other and there are only two outcomes

X:Bin(39, 0.12)

Probability of finding five left-handers

=P(X=5)

= 39C5 (0.12)^5 (0.88)^34

=0.1856

Answer is 0.1856