The probability that 15 or more students in the sample are left-handed handed is 0.0752
(a) Why L can be a binomial distribution
From the question, we understand that 11% of the students are left-handed; this means that 89% are not left-handed
This, in other words means that:
For each observation, the student is either left-handed or not left-handed
Hence, the distribution can be represented by a binomial distribution
(b) The probability that 15 or more are left-handed
The sample size is given as:
[tex]\mathbf{n = 100}[/tex]
And the proportion is
[tex]\mathbf{p = 11\%}[/tex]
The mean of the dataset is:
[tex]\mathbf{\mu = np}[/tex]
[tex]\mathbf{\mu = 100 \times 11\%}[/tex]
[tex]\mathbf{\mu = 11}[/tex]
The standard deviation of the dataset is:
[tex]\mathbf{\sigma = \sqrt{\mu(1- p)}}[/tex]
So, we have:
[tex]\mathbf{\sigma = \sqrt{11 \times (1- 11\%)}}[/tex]
[tex]\mathbf{\sigma = \sqrt{11 \times 89\%}}[/tex]
[tex]\mathbf{\sigma = \sqrt{9.79}}[/tex]
[tex]\mathbf{\sigma = 3.13}[/tex]
So, the probability is represented as:
[tex]\mathbf{P(x \ge 15)}[/tex]
Using continuity correction, we have:
[tex]\mathbf{P(x \ge 15) = P(x \ge 15.5)}[/tex]
Calculate the z-score for x = 15.5
[tex]\mathbf{z =\frac{x - \mu}{\sigma}}[/tex]
So, we have:
[tex]\mathbf{z =\frac{15.5 - 11}{3.13}}[/tex]
[tex]\mathbf{z =\frac{4.5}{3.13}}[/tex]
[tex]\mathbf{z =1.438}[/tex]
The probability is then calculated as:
[tex]\mathbf{P(x \ge 15) = P(z \ge 1.438)}[/tex]
Using z-tables of probabilities, we have:
[tex]\mathbf{P(x \ge 15) = 0.075217}[/tex]
Approximate
[tex]\mathbf{P(x \ge 15) = 0.0752}[/tex]
Hence, the probability is 0.0752
Read more about binomial distribution at:
https://brainly.com/question/15992180
