East LA College M227 Statistics: Exam #4 According to Nielsen Media Research, 25

Question

East LA College M227 Statistics: Exam #4 According to Nielsen Media Research, 25% of TV-s are tuned to NFL Monday Night Football when it is televised. Assuming the show is being broadcast and that 30 TV-s are randomly selected 1. Determine the mean, variance, and standard deviation for the distribution 2. Determine the probability that 5 [out of 30 TV-s] are tuned to NFL Monday Night Football 3. Less than 10 (out of 30 TV-s] are tuned to NFL Monday Night Football . Find the most expected number of TV-s [out of 30 TV-s], that are tuned to NFL Monday Night Football

Explanation / Answer

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
a.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 30 * 0.25
= 7.5
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 30 * 0.25 * 0.75
= 5.625
III.
standard deviation = sqrt( variance ) = sqrt(5.625)
=2.371708
b.
P( X = 5 ) = ( 30 5 ) * ( 0.25^5) * ( 1 - 0.25 )^25
= 0.104728
c.
P( X < 10) = P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 30 9 ) * 0.25^9 * ( 1- 0.25 ) ^21 + ( 30 8 ) * 0.25^8 * ( 1- 0.25 ) ^22 + ( 30 7 ) * 0.25^7 * ( 1- 0.25 ) ^23 + ( 30 6 ) * 0.25^6 * ( 1- 0.25 ) ^24 + ( 30 5 ) * 0.25^5 * ( 1- 0.25 ) ^25 + ( 30 4 ) * 0.25^4 * ( 1- 0.25 ) ^26 + ( 30 3 ) * 0.25^3 * ( 1- 0.25 ) ^27 + ( 30 2 ) * 0.25^2 * ( 1- 0.25 ) ^28 + ( 30 1 ) * 0.25^1 * ( 1- 0.25 ) ^29 + ( 30 0 ) * 0.25^0 * ( 1- 0.25 ) ^30
= 0.803407
d.
P( X < 20) = ( 30 19 ) * 0.25^19 * ( 1- 0.25 ) ^11 + ( 30 18 ) * 0.25^18 * ( 1- 0.25 ) ^12 + ( 30 17 ) * 0.25^17 * ( 1- 0.25 ) ^13 + ( 30 16 ) * 0.25^16 * ( 1- 0.25 ) ^14 + ( 30 15 ) * 0.25^15 * ( 1- 0.25 ) ^15 + ( 30 14 ) * 0.25^14 * ( 1- 0.25 ) ^16 + ( 30 13 ) * 0.25^13 * ( 1- 0.25 ) ^17 + ( 30 12 ) * 0.25^12 * ( 1- 0.25 ) ^18 + ( 30 11 ) * 0.25^11 * ( 1- 0.25 ) ^19 + ( 30 10 ) * 0.25^10 * ( 1- 0.25 ) ^20 +
= 0.999998

this can be approximate to value 20 out of 30

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.