Mean and standard deviation of a binomial variable

About this lesson

Statistics (AP-aligned) › Random variables

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Mean and standard deviation of a binomial variable

Warm-up

Three quick questions from earlier lessons.

Discover

A fair coin is tossed 100 times. About how many heads do you expect?

  1. 50
  2. 25

Key idea

Shortcuts for the binomial.

If XX is binomial with nn and pp: μX=np\mu_X = np and σX=np(1−p)\sigma_X = \sqrt{np(1 - p)}.

Worked example

n=50n = 50, p=0.3p = 0.3.

μ=15\mu = 15
np.
σ2=10.5\sigma^2 = 10.5
np(1 − p).
σ≈3.24\sigma \approx 3.24
Square root.

Your turn, guided

Your turn: n=200n = 200, p=0.1p = 0.1. The mean.

2020
200 × 0.1.

Practice ★

  1. X is binomial with n = 91 and p = 1/9. Find E(X).
  2. X is binomial with n = 87 and p = 1/8. Find E(X).
  1. X is binomial with n = 70 and p = 1/5. Find SD(X).
  2. X is binomial with n = 71 and p = 1/2. Find SD(X).

Watch out!

Practice ★★

  1. A multiple-choice test has 87 questions with 4 options each. A student guesses every answer. What does E(X) = \frac{87}{4} mean?
    1. Over many such guessed tests, the average number correct is about \frac{87}{4}
    2. The student will get exactly \frac{87}{4} right
    3. \frac{87}{4} is the largest possible score

Exit ticket

Three questions to check you have got it.

  1. X is binomial with n = 91 and p = 1/9. Find E(X).
  2. X is binomial with n = 87 and p = 1/8. Find E(X).
  3. X is binomial with n = 59 and p = 1/5. Find E(X).

Summary card

Binomial mean and SD

μ=np\mu = np, σ=np(1−p)\sigma = \sqrt{np(1 - p)}.

Watch out: the SD is the square root of np(1−p)np(1 - p).

End of the lesson

Next lesson: The geometric distribution Back to the chapter

Answers: Mean and standard deviation of a binomial variable

A

  1. 919\frac{91}{9}
  2. 878\frac{87}{8}

B

  1. 2705\frac{2 \sqrt{70}}{5}
  2. 712\frac{\sqrt{71}}{2}

C

  1. Over many such guessed tests, the average number correct is about \frac{87}{4}

D

  1. 919\frac{91}{9}
  2. 878\frac{87}{8}
  3. 595\frac{59}{5}

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