Question

Il bullom 20% of all middle-aged men? 14 I preference survey showed that 60% of all households are watching a sports-related program. Lets assume this figure is true. Suppose a random sample of 10 households is contacted. Let X = the number of households watching a sports-related program. a) Explain briefly why X can be considered a binomial random variable. (3 pts) b) Explain what P(X= 3) means in words, and find its value. (3 pts) c) Find the probability that at least 7 households are watching a sports-related program. (3 pts) d) Find the probability that from 2 to 5 households (including 2 and 5) are watching a sports related program. (3 pts)

Answer 1

(a) 'X' as denoted is the number of households watching sports.

  Out of 10 samples, there can be only things either the household watches sports or they don't. Since there are only 2 possible outcomes and more n > 1, we can say that 'X' is a binomial random variable.

Binomial is an experiment of 'n' trials where there can be only 2 outcomes -success and failure. (success is denoted by the variable and failure is the opposite.) Success and failure is defined differently for different experiments.)

X Bin(10,0.60)

P(X = 1) = nCr *p"*(1-pn-1

Where

n! nCr=7 (n- )!!

(b)  P (X = 3) is the probability that exactly 3 out of 10 households watch Sports.

  
P(X = 3) = 10C3 * 0.60%*(1 -0.6010-3

=

0.0425

(c) Probability that at least 7 households watch sports mean 7 or more than 7

P(X >7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

=
10C7 * 0.60' (1 -0.60)10-7+ .........10C10 * 0.600

= 0.2150 + 0.1209 + 0.0403 + 0.0060

=

03823

(d) Probability that 2 - 5 households watch sports .

P(2 < X <5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

=

10C2 * 0.602 + (1 -0.60)10-2 + .........1005 *0.60*(1 -0.60)10-5

= 0.106 + 0.0425 + 0.1115 + 0.2007

=

03652

(e) The expected value of binomial = np

E (X) = 10 * 0.60

E(X) = 6