Question

A total of 11% of students at a large high school are left-handed. A statistics teacher selects a random sample of 100 students ; and records L = the number of left-handed students in the sample. What is the probability that 15 or more students in the sample are left-handed?

A) 0.867
B) 0.0802
C) 0.0528
D) 0.9198
E) 0.133

Answer 1

Explanation:

Number of students in the sample, n = 100

Percentage of left-handed students, p = 0.11

The mean (expected value) of the sample is:

μ=n⋅p=100⋅0.11=11

The standard deviation of the sample is:

σ=n⋅p⋅(1−p)​=100⋅0.11⋅0.89​≈3.14

Now, let’s find the z-score for 15 left-handed students:

Z=15−μ/σ ​=15−11/3.14​≈1.28

Using the standard normal distribution table or calculator, we find:

P(X≥15)=P(Z≥1.28)≈1−0.8997=0.1003

Therefore, the probability that 15 or more students in the sample are left-handed is approximately 0.1003