Compute PiX) using the binomial probability formula. Then determine whelher the

Question

Compute PiX) using the binomial probability formula. Then determine whelher the nomal distributon can be used to estimate his probability. IT so, appreximate PiX) using the nermal distribution and compare the result wilh the exact probability n = 30, p = 0.35, and X = 15 PXRound to four decimal places as needed.) Can the normal distribution be usod to approximate this probablity? A. No the norrnal distribution cannol be used because np(1-p)-10. O B. Yes, the normal distnbution can be used because npl p)10 O C. No the normal distribution cannot he used because np(1-p)s 10 0 D. Yes, the normal distribution can be used because no(1-p)-:10. Approximate PIx) using the normal distibution. Usc a standard ncrmal distribution table. Scloct the correct choice below and til in any answer boxes in your choice Round to tour decimal place as needed.] B. There is no solution By how much do the exact and approximated probabilities differ? Select the correct choice below and fill in any answer boxes in your choice °A.- (Round to for decimal paces as needed) O B. There is no solution

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
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 30 * 0.35
= 10.5
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 30 * 0.35 * 0.65
= 6.825
III.
standard deviation = sqrt( variance ) = sqrt(6.825)
=2.61247
P( X = 15 ) = ( 30 15 ) * ( 0.35^15) * ( 1 - 0.35 )^15
= 0.03511
a.
yes,normal ditribution can used because np(1-p)>10

b.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 30 * 0.35 = 10.5
standard deviation ( npq )= 30*0.35*0.65 = 2.6125
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
P(X > 15) = (15-10.5)/2.6125
= 4.5/2.6125 = 1.7225
= P ( Z >1.7225) From Standard Normal Table
= 0.0425
P(X < = 15) = (1 - P(X > 15))
= 1 - 0.9575 = 0.0425

c.
p(X=15)=0.03511,p(x<=15)=0.0425
both they differ,
0.0425-0.03511 = 0.0073

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