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:
[tex] P(x) = CC\left(\begin{array}{ccc}n\\x\end{array}\right) p^x q^(^n^-^x^) [/tex]
Thus,
[tex]P(12) = C\left(\begin{array}{ccc}20\\12\end{array}\right) (0.7^1^2) (0.3^(^2^0^-^1^2^) )[/tex]
[tex] = 125970*0.0138*0.00006 [/tex]
[tex] = 0.1144 [/tex]
b) [tex]P(16) = C\left(\begin{array}{ccc}20\\16\end{array}\right) 0.7^1^6 (0.3^(^2^0^-^1^6^))[/tex]
[tex] = 4845 * 0.0033 * 0.0081 [/tex]
[tex] = 0.1304 [/tex]
c) Compute P(x≥16):
P(x ≥ 16) = P(16) + P(17) + P(18) + P(19) + P(20)
[tex]= 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^))[/tex]
[tex] = 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.2375 [/tex]
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
