Question

question 1 of 9
eleven percent of students at a large high school are left - handed. a statistics teacher selects a random sample of 100 students and records (x=) the number of left - handed students in the sample.
is the probability distribution of (x) approximately normal? justify your answer.
(np=) (square) and (n(1 - p)=) (square)
therefore, the probability distribution of (x) approximately normal.

Explanation:

Step1: Identify n and p

We know that the sample size \( n = 100 \) (since a random sample of 100 students is selected) and the probability of a student being left - handed \( p=0.11 \) (eleven percent of students are left - handed).

Step2: Calculate \( np \)

To calculate \( np \), we multiply the sample size \( n \) by the probability \( p \). So, \( np=100\times0.11 = 11 \).

Step3: Calculate \( n(1 - p) \)

First, calculate \( 1-p=1 - 0.11 = 0.89 \). Then, multiply \( n \) by \( (1 - p) \). So, \( n(1 - p)=100\times0.89 = 89 \).

Step4: Check normal approximation conditions

For a binomial distribution \( X\sim B(n,p) \) to be approximately normal, we need \( np\geq10 \) and \( n(1 - p)\geq10 \). Since \( np = 11\geq10 \) and \( n(1 - p)=89\geq10 \), the probability distribution of \( X \) is approximately normal.

Answer:

\( np = 11 \) and \( n(1 - p)=89 \). Since \( np\geq10 \) and \( n(1 - p)\geq10 \), the probability distribution of \( X \) is approximately normal.