Binomial Distribution

Question

fit a binomial distribution to the following data ?

x 0 1 2 3 4 5 6 Total

f 5 18 28 12 7 6 4 80

Answer 1

x f x*f

0 5 0

1 18 18

2 28 56

3 12 36

4 7 28

5 6 30

6 4 24

80 192

N = 80

mean of frequency distribution is sum xf / sum f = 2.4

mean of binomial distribution is np = 2.4

n here is 6, so 6p = 2.4

p = 0.4

q = 1 - p = 1 - 0.4 = 0.6

r = 0, 1, 2, 3, 4, 5, 6

Per binomial probability formula:

P(r) = C(n,r) * p^r * q^(n-r), where C(n,r) = n!/(r! n-r!)

P(r = 0) = C(6,0) (0.4)^0 (0.6)^6 = 0.047

P(r = 1) = C(6,1) (0.4)^1 (0.6)^5 = 0.187

P(r = 2) = C(6,2) (0.4)^2 (0.6)^4 = 0.311

P(r = 3) = C(6,3) (0.4)^3 (0.6)^3 = 0.276

P(r = 4) = C(6,4) (0.4)^4 (0.6)^2 = 0.138

P(r = 5) = C(6,5) (0.4)^5 (0.6)^1 = 0.037

P(r = 6) = C(6,6) (0.4)^6 (0.6)^0 = 0.004

so

Binomial Distribution

r P(r) Expected Frequency N * P(r), where N is 80

0 0.047 3.76 ~ 4

1 0.187 14.96 ~ 15

2 0.311 24.88 ~25

3 0.276 22.80 ~23

4 0.138 11.04 ~11

5 0.037 2.96 ~3

6 0.004 0.032 0

81 (round off error)