Question

Consider a binomial experiment with n = 20 and p = .70.

Answer 1

Complete question:

Consider a binomial experiment with n = 20 and p = .70.

A. Compute f(12).

B. Compute f(16).

C. Compute P(x≥ 16).

D. Compute P(x≤15).

E. Compute E(x).

F. Compute Var(x).

Answer:

a) 0.1144

b) 0.1304

c) 0.2375

d) 0.7625

e) 14

f) 4.2

Step-by-step explanation:

Given:

n = 20

p = 0.70

q = 1 - p ==>  1 - 0.70 = 0.30

a) Use the formula:

$P(x) = CC\left(\begin{array}{ccc}n\\x\end{array}\right) p^x q^(^n^-^x^)$

Thus,

$P(12) = C\left(\begin{array}{ccc}20\\12\end{array}\right) (0.7^1^2) (0.3^(^2^0^-^1^2^) )$

$= 125970*0.0138*0.00006$

$= 0.1144$

b) $P(16) = C\left(\begin{array}{ccc}20\\16\end{array}\right) 0.7^1^6 (0.3^(^2^0^-^1^6^))$

$= 4845 * 0.0033 * 0.0081$

$= 0.1304$

c) Compute P(x≥16):

P(x ≥ 16) = P(16) + P(17) + P(18) + P(19) + P(20)

$= C\left(\begin{array}{ccc}20\\16\end{array}\right) 0.7^1^6 (0.3^(^2^0^-^1^6^)) + C\left(\begin{array}{ccc}20\\17\end{array}\right) 0.7^1^7 (0.3^(^2^0^-^1^7^) ) + C\left(\begin{array}{ccc}20\\18\end{array}\right) 0.7^1^8 (0.3^(^2^0^-^1^8^)) + C\left(\begin{array}{ccc}20\\19\end{array}\right) 0.7^1^9 (0.3^(^2^0^-^1^9^)) + C\left(\begin{array}{ccc}20\\20\end{array}\right) 0.7^2^0 (0.3^(^2^0^-^2^0^))$

$= 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.2375$

d) P(x ≤ 15):

= 1 - P(x ≥ 16)

= 1 - 0.2375

= 0.7625

e) E(x): use the formula, n * p.

= n*p

= 20 * 0.7

= 14

f) Var(x)

Use the formula: npq

npq = 20 * 0.7 * 0.3

= 4.2

σ