A study shows that 10% of americans are left-handed.
Find the probability that at most one student is left handed in a class of 20 students.
P(x)=.10
"at most" means "binomialCDF" on calculator.
1 student is upper value, 0 is lower value.
n=20 or number of trials.
ANSWER: CDF=.39 or ~39% that at most one student is left handed in a class of 20 students.
To find the probability that at most one student is left-handed in a class of 20 students, we can use the binomial probability formula.
The binomial probability formula is given by:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting exactly x successes
C(n, x) is the number of combinations of n things taken x at a time
p is the probability of success for each individual trial
n is the number of trials
In this case, the probability of an individual being left-handed is 10% or 0.1, so p = 0.1. The number of trials is 20, so n = 20.
We need to calculate the probability of at most one student being left-handed, which means we need to calculate the probabilities of 0 and 1 students being left-handed and sum them up.
P(0) = C(20, 0) * 0.1^0 * (1-0.1)^(20-0)
P(0) = 1 * 1 * 0.9^20
P(0) ≈ 0.1216 (rounded to four decimal places)
P(1) = C(20, 1) * 0.1^1 * (1-0.1)^(20-1)
P(1) = 20 * 0.1 * 0.9^19
P(1) ≈ 0.2702 (rounded to four decimal places)
Now we can sum up the probabilities:
P(at most one student is left-handed) = P(0) + P(1)
P(at most one student is left-handed) ≈ 0.1216 + 0.2702
P(at most one student is left-handed) ≈ 0.3918 (rounded to four decimal places)
Therefore, the probability that at most one student is left-handed in a class of 20 students is approximately 0.3918 or 39.18%.