Question

Find the? mean, variance, and standard deviation of the binomial distribution with the given values of n and p.

n equals 122n=122?,

p equals 0.62p=0.62

The? mean,

mu??,

is

75.675.6.

?(Round to the nearest tenth as? needed.) The? variance,

sigma?squared2?,

is

nothing.

?(Round to the nearest tenth as? needed.) The standard? deviation,

sigma??,

is

nothing.

?(Round to the nearest tenth as? needed.

Answer 1

Given,

n = 122, p = 0.62

For binomial distribution,

Mean = np

= 122 * 0.62

= 75.6

Variance ² = n p ( 1 - p)

= 122 * 0.62 * 0.38

= 28.7

² = 28.7

Standard deviation = Sqrt ( Variance )

= Sqrt( n p(1-p) )

= 5.4

= 5.4