Suppose that you are in a class of 28 students and it is assumed that approximately 8% of the population is left-handed. (Give your answers correct to three decimal place.
Compute the probability that exactly five students are left-handed
Pr=.8^5*.2^20 assuming students are either left OR right handed, and none are ambidextrous.
To compute the probability that exactly five students are left-handed in a class of 28 students, we can use the binomial probability formula. The binomial probability formula is given by:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of having exactly k successes
- n is the total number of trials
- k is the number of successful trials
- p is the probability of success in a single trial
- (n choose k) is the binomial coefficient, which can be calculated as n! / (k! * (n-k)!)
In this case, we have:
- n = 28 (total number of students)
- k = 5 (number of left-handed students)
- p = 0.08 (probability of a student being left-handed)
Substituting these values into the formula, we get:
P(X = 5) = (28 choose 5) * 0.08^5 * (1-0.08)^(28-5)
Now, let's calculate each component step-by-step:
1. Calculating the binomial coefficient (28 choose 5):
(28 choose 5) = 28! / (5! * (28-5)!)
= 28! / (5! * 23!)
= (28 * 27 * 26 * 25 * 24) / (5 * 4 * 3 * 2 * 1)
= 98,280
2. Calculating p^k * (1-p)^(n-k):
0.08^5 * (1-0.08)^(28-5) = 0.08^5 * 0.92^23
≈ 0.0024
3. Combining the results:
P(X = 5) ≈ (28 choose 5) * 0.08^5 * (1-0.08)^(28-5)
≈ 98,280 * 0.0024
≈ 235.872
Therefore, the probability that exactly five students are left-handed in a class of 28 students is approximately 0.236.