"In a local college, 20% of the math..." was answered by our community

Vagotonusybk

Answered question

2022-02-11

In a local college,

20 %

of the math majors are women. Ten math majors are chosen at random.
a. What is the probability that exactly 2 are women?
b. How do you find the mean and the variance?

Answer & Explanation

Clark Carson

Beginner2022-02-12Added 17 answers

Step 1
$P (X = 2), n = 10, p = 0.2$
$X \sim Binomial (n = 10, p = 0.2)$
$P (X = 2) = (\begin{matrix} n \\ x \end{matrix}) p^{x} {(1 - p)}^{n - x} = (\begin{matrix} 10 \\ 2 \end{matrix}) {(0.2)}^{2} {(1 - 0.2)}^{10 - 2}$
$= 45 \times {(0.2)}^{2} {(0.8)}^{8}$
$= 0.30199 \approx 0.301990 \approx 0.302$
$P (X = 2) = 0.302$
Step 2
$P (X \leq 2), n = 10, p = 0.2$
$(X \leq 2) = P (0 \leq X \leq 2) = P (X = 0, 1, 2)$
$= P (X = 0) + P (X = 1) + P (X = 2)$ (because these are disjoint events)
$P (X = 0) = (\begin{matrix} 10 \\ 0 \end{matrix}) {(0.2)}^{0} {(1 - 0.2)}^{10} = 1 {(0.2)}^{0} {(0.8)}^{10} \approx 0.107374$
$P (X = 1) = (\begin{matrix} 10 \\ 1 \end{matrix}) {(0.2)}^{1} {(1 - 0.2)}^{9} = 10 {(0.2)}^{1} {(0.8)}^{9} =\approx 0.268435$
$P (X = 2) = (\begin{matrix} 10 \\ 2 \end{matrix}) {(0.2)}^{2} {(1 - 0.2)}^{8} = 45 {(0.2)}^{2} {(0.8)}^{8} \approx 0.30199$
$P (X = 0) + P (X = 1) + P (X = 2)$
$= 0.107374 + 0.268435 + 0.30199 = 0.677799 \approx 0.678$
$P (X \leq 2) = 0.678$
Step 3
$P (X = 0), n = 10, p = 0.2$
$X \sim Binomial (n = 10, p = 0.2)$
$<$

Johnathan Carpenter

Beginner2022-02-13Added 12 answers

Step 1
b) Mean,

μ = n \times p = 10 \times 0.2 = 2

Variance,

s = n \times p \times (1 - p) = 10 \times 0.2 \times (1 - 0.2) = 1.6

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