Question

According to a survey of 25,000 households, 47% of the people watching a special event on TV enjoyed the commercials more than the event itself. Complete parts a through e below based on a random sample of nine people who watched the special event.

a. What is the probability that exactly three people enjoyed the commercials more than the? event?

b. What is the probability that less than four people enjoyed enjoyed the commercials more than the event ?

c.  What is the probability that exactly five or six people enjoyed the commericals more than the event?

d. What are the mean and standard deviation for this distribution?

Please explain

Answer 1

Solution:

p = 0.47

q = 1 - p = 1- 0.47 = 0.53

n = 9

binomial probability distribution

Formula:

P_{(k out of n )}= n!*p^k * q^n-k / k! *(n - k)!

( a )

P( x = 3 ) = 9!*0.47³ * 0.53^9-3 / 3! *(9 - 3)! = 0.1933

( b )

P( x < 4 ) = 9!*0.47³ * 0.53^9-3 / 3! *(9 - 3)! +  9!*0.47² * 0.53^9-2 / 2! *(9 - 2)! + 9!*0.47¹ * 0.53^9-1 / 1! *(9 - 1)!

+  9!*0.47⁰ * 0.53^9-0 / 0! *(9 - 0)!

= 0.1993+0.0934+0.0263+0.0033

= 0.3164  

( c )

P( x = 5 or 6 ) =  9!*0.47⁵ * 0.53^9-5 / 5! *(9 - 5)! + 9!*0.47⁶ * 0.53^9-6 / 6! *(9 - 6)!

= 0.2280+0.1348

= 0. 3628

( d )

x	p( x )
0	0.0033
1	0.0263
2	0.0934
3	0.1933
4	0.2571
5	0.2280
6	0.1348
7	0.0512
8	0.0114
9	0.0012

Mean = = x . p( x )

= 0*0.0033+1*0.0263+2*0.0934+3*0.1933+4*0.2571+5*0.2280+6*0.1348+7*0.0512+8*0.0114+9*0.0012

= 4.2306

Standard deviation = =

= 0²*0.0033+1²*0.0263+2²*0.0934+3²*0.1933+4²*0.2571+5²*0.2280+6²*0.1348+7²*0.0512+8²*0.0114+9²*0.0012

= 20.1416

= = = 1.4979