Question

If x is binomially distributed with p=.62 find the following probabilities and round to 4 decimal places
(a) P(11) n=22
(b) P(16) n=25
(c) P(20) n=29
(d) P(8) n=12
(e) P(x>16) n=24
(f) P(x_>11) n=17
(g) P(8_ (h) P(14

Answer 1

Probabilities are used to determine the chances of events

The values of the binomial probabilities are

• P(11) = 0.0876
• P(16) =0.1609
• P(20) =0.1166
• P(8) = 0.2254
• P(x > 16) = 0.2513
• P(x >= 11) = 0.5161

How to evaluate the binomial probabilities

The proportion (p) is given as:

p = 0.62

The binomial probability is represented as:

$P(x) = ^nC_x * p^x * (1 - p)^{n -x}$

Using the above equation, we have:

(a) P(11) n=22

$P(11) = ^{22}C_{11} * 0.62^{11} * (1 - 0.62)^{22 -11}$

$P(11) = 0.0876$

(b) P(16) n=25

$P(16) = ^{25}C_{16} * 0.62^{16} * (1 - 0.62)^{25 -16}$

$P(16) =0.1609$

(c) P(20) n=29

$P(20) = ^{29}C_{20} * 0.62^{20} * (1 - 0.62)^{29 -20}$

$P(20) =0.1166$

(d) P(8) n=12

$P(8) = ^{12}C_{8} * 0.62^{8} * (1 - 0.62)^{12 -8$

$P(8) = 0.2254$

(e) P(x>16) n=24

$P(x > 16) = P(17) + P(18)+....+P(24)$

$P(x > 16) = 0.2513$

(f) P(x_>11) n=17

$P(x \ge 11) = P(11) + P(12)+....+P(17)$

$P(x \ge 11) = 0.5161$

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