Estimate the binomial distribution using the standard normal approximation. Draw

Question

Estimate the binomial distribution using the standard normal approximation. Draw the picture. Lakewood lumber orders windows from a prominent window manufacturer. The window company has estimated that 3% of the windows will have locks that look improperly and are considered defective. Assume a production run of 175 windows are completed a) what is the mean and standard deviation of this distribution b) What is the probability that 6 or more are defective c) what is the probability that exactly 6 windows are defective d) what is the probability that there are more than 8 defective windows

Explanation / Answer

Binomial 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.
Mean ( np ) =175 * 0.03 = 5.25
Standard Deviation ( npq )= 175*0.03*0.97 = 2.2567

b.
P( X < 6) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 175 5 ) * 0.03^5 * ( 1- 0.03 ) ^170 + ( 175 4 ) * 0.03^4 * ( 1- 0.03 ) ^171 + ( 175 3 ) * 0.03^3 * ( 1- 0.03 ) ^172 + ( 175 2 ) * 0.03^2 * ( 1- 0.03 ) ^173 + ( 175 1 ) * 0.03^1 * ( 1- 0.03 ) ^174 + ( 175 0 ) * 0.03^0 * ( 1- 0.03 ) ^175
= 0.5715
P( X > = 6 ) = 1 - P( X < 6) = 0.4285
c.
P( X = 6 ) = ( 175 6 ) * ( 0.03^6) * ( 1 - 0.03 )^169
= 0.155
d.P( X < = 8) = 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)   
= ( 175 8 ) * 0.03^8 * ( 1- 0.03 ) ^167 + ( 175 7 ) * 0.03^7 * ( 1- 0.03 ) ^168 + ( 175 6 ) * 0.03^6 * ( 1- 0.03 ) ^169 + ( 175 5 ) * 0.03^5 * ( 1- 0.03 ) ^170 + ( 175 4 ) * 0.03^4 * ( 1- 0.03 ) ^171 + ( 175 3 ) * 0.03^3 * ( 1- 0.03 ) ^172 + ( 175 2 ) * 0.03^2 * ( 1- 0.03 ) ^173 + ( 175 1 ) * 0.03^1 * ( 1- 0.03 ) ^174 + ( 175 0 ) * 0.03^0 * ( 1- 0.03 ) ^175   
= 0.9175
P( X > 8) = 1 - P ( X <=8) = 1 -0.9175 = 0.0825

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