Question

Suppose x is a binomial random variable with p = .4 and n = 25.

c. Use the binomial probabilities table or statistical software to find the exact value of P(x>=9). Answ:.726 back of book

d. Use the normal approximation to find P(x>=9). answ:.7291 the back of book

For one I have no idea how to use the binomial probabilities table .

The mean is 10, variance is 6 and std is 2.45

If possible could someone explain how to get those answers and how to use the binomial probabilities table. Not just the math but actually explain how you solve because I have no idea how and really need help

Answer 1

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
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 25 * 0.4
= 10
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 25 * 0.4 * 0.6
= 6
III.
standard deviation = sqrt( variance ) = sqrt(6)
=2.4495

a.
mean is 10
b.
variance =6
standard deviation =2.45
c.
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)   
= ( 25 8 ) * 0.4^8 * ( 1- 0.4 ) ^17 + ( 25 7 ) * 0.4^7 * ( 1- 0.4 ) ^18 + ( 25 6 ) * 0.4^6 * ( 1- 0.4 ) ^19 + ( 25 5 ) * 0.4^5 * ( 1- 0.4 ) ^20 + ( 25 4 ) * 0.4^4 * ( 1- 0.4 ) ^21 + ( 25 3 ) * 0.4^3 * ( 1- 0.4 ) ^22 + ( 25 2 ) * 0.4^2 * ( 1- 0.4 ) ^23 + ( 25 1 ) * 0.4^1 * ( 1- 0.4 ) ^24 + ( 25 0 ) * 0.4^0 * ( 1- 0.4 ) ^25
= 0.2735
P( X > = 9 ) = 1 - P( X < 9) = 0.7265

d.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 25 * 0.4 = 10
standard deviation ( √npq )= √25*0.4*0.6 = 2.4495
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
P(X < 9) = (9-10)/2.4495
= -1/2.4495= -0.4082
= P ( Z <-0.4082) From Standard NOrmal Table
= 0.3415
P(X > = 9) = (1 - P(X < 9))
= 1 - 0.3415 = 0.6585