A quiz has 20 multiple-choice questions with four choices each, and a student guesses every answer at random. Let X be the number of correct answers. Find the expected value, variance and standard deviation of X, and the probability of exactly 5 correct.
1
Check that the count is binomial
There are n=20 trials, each guess is right or wrong, the guesses are independent, and each is right with probability p=1/4=0.25.X is therefore binomial.
2
Expected value
Multiply the number of trials by the chance of success on each one.
X∼B(20,0.25)E(X)=np=20⋅0.25=5
3
Variance and standard deviation
The chance of a wrong guess is 1−p=0.75. The standard deviation is the square root of the variance.
Choose which 5 of the 20 guesses are right, C(20,5)=15,504 ways, and multiply by the probability of one such arrangement. A calculator gives about 0.2023.
A guesser averages 5 correct, and about 93% of the time the score is within two standard deviations of that. Those bounds are 5±2×1.94, about 1.1 to 8.9, so the scores 2through 8. Exactly 5 is the single most likely score, yet it turns up only about one time in five. Ranges come from adding entries: P(4≤X≤6)=0.1897+0.2023+0.1686=0.5606. A score of 10 or more is rare: adding P(X=k) from k=10 to 20 gives about 0.014.
Result
E(X)=5,Var(X)=3.75,SD(X)≈1.94 and P(X=5)≈0.2023.
Your turn
1. One like the example
X is binomial with n=50 and p=0.4. Find E(X),Var(X) and SD(X).
Show the answer and explanation
E(X)=20,Var(X)=12 and SD(X)≈3.46.
E(X)=np=50⋅0.4=20.Var(X)=np(1−p)=50⋅0.4⋅0.6=12, and SD(X)=√12≈3.46.
A player makes 80% of her free throws, independently. She takes 10. What is the probability she makes exactly 8?
Show the answer and explanation
About 0.3020.
X∼B(10,0.8) and k=8, so two shots miss. C(10,8)=45 arrangements, each with probability (0.8)8(0.2)2. The expected count is 8, and 8 is also the single most likely count, at about 30%.
A vaccine gives a mild fever to 5% of people, independently. A clinic vaccinates 40 people and 6 get a fever. Find the expected count and its standard deviation. Is 6 unusual?
Show the answer and explanation
E(X)=2 and SD(X)≈1.38. Six is about 2.9 standard deviations above the expected count, which is unusual.
X∼B(40,0.05), so E(X)=2 and Var(X)=40⋅0.05⋅0.95=1.9. The z-like distance (6−2)/1.38≈2.9 is well past 2. The exact chance of 6 or more is about 0.014, so the clinic might check its batch.
Open the rows in Math and change the variance row to Var(X)=np=20⋅0.25=5, leaving out 1−p. The row turns to ✗: “Doesn’t match Var(X) = np(1 − p) for a binomial.” Tap the ✗ and choose Show the value to see what the rule gives.
