determined effort to Before 1918, approximately 60% of the wolves in the New Mex

Question

determined effort to Before 1918, approximately 60% of the wolves in the New Mexico and Anona region were male, and 40% were female. However, eattle exterminate wolves. From 1918 to the presert, approximately 70% er wolves in the reg on re male. 30% are femaleheegia esped pet me rees re moe eer ee an area where the population has been greatly reduced. (Round your answers to three decimal places ranchers in this area have made (a) Before 1918, in a random sample of 10 wolves spotted in the region, what is the probability that 7 or more were male? What is the probability that 7 or more were female? What is the probability thet fewer than 4 were female? (b) For the period from 1918 to the present, in a random sample of 10 wolves spot ted in the region, what is the probabilty that 7 or more were male What is the probability that 7 or more were female? What is the probability that fewer than 4 were female Need Help? ToTr

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 = 10 * 0.6
= 6
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 10 * 0.6 * 0.4
= 2.4
III.
standard deviation = sqrt( variance ) = sqrt(2.4)
=1.54919
a.
P( X < 7) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 10 6 ) * 0.6^6 * ( 1- 0.6 ) ^4 + ( 10 5 ) * 0.6^5 * ( 1- 0.6 ) ^5 + ( 10 4 ) * 0.6^4 * ( 1- 0.6 ) ^6 + ( 10 3 ) * 0.6^3 * ( 1- 0.6 ) ^7 + ( 10 2 ) * 0.6^2 * ( 1- 0.6 ) ^8 + ( 10 1 ) * 0.6^1 * ( 1- 0.6 ) ^9 + ( 10 0 ) * 0.6^0 * ( 1- 0.6 ) ^10
= 0.61772
P( X > = 7 ) = 1 - P( X < 7) = 0.38228
b.
P( X < 7) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 10 6 ) * 0.4^6 * ( 1- 0.4 ) ^4 + ( 10 5 ) * 0.4^5 * ( 1- 0.4 ) ^5 + ( 10 4 ) * 0.4^4 * ( 1- 0.4 ) ^6 + ( 10 3 ) * 0.4^3 * ( 1- 0.+
= 0.94524
P( X > = 7 ) = 1 - P( X < 7) = 0.05476
c.
P( X < 4) = P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 10 3 ) * 0.4^3 * ( 1- 0.4 ) ^7 + ( 10 2 ) * 0.4^2 * ( 1- 0.4 ) ^8 + ( 10 1 ) * 0.4^1 * ( 1- 0.4 ) ^9 + ( 10 0 ) * 0.4^0 * ( 1- 0.4 ) ^10
= 0.38228

PART B.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 10 * 0.7
= 7
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 10 * 0.7 * 0.3
= 2.1
III.
standard deviation = sqrt( variance ) = sqrt(2.1)
=1.44914

P( X < 7) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 10 6 ) * 0.7^6 * ( 1- 0.7 ) ^4 + ( 10 5 ) * 0.7^5 * ( 1- 0.7 ) ^5 + ( 10 4 ) * 0.7^4 * ( 1- 0.7 ) ^6 + ( 10 3 ) * 0.7^3 * ( 1- 0.7 ) ^7 + ( 10 2 ) * 0.7^2 * ( 1- 0.7 ) ^8 + ( 10 1 ) * 0.7^1 * ( 1- 0.7 ) ^9 + ( 10 0 ) * 0.7^0 * ( 1- 0.7 ) ^10
= 0.35039
P( X > = 7 ) = 1 - P( X < 7) = 0.64961

P( X < 7) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 10 6 ) * 0.3^6 * ( 1- 0.3 ) ^4 + ( 10 5 ) * 0.3^5 * ( 1- 0.3 ) ^5 + ( 10 4 ) * 0.3^4 * ( 1- 0.3 ) ^6 + ( 10 3 ) * 0.3^3 * ( 1- 0.3 ) ^7 + ( 10 2 ) * 0.3^2 * ( 1- 0.3 ) ^8 + ( 10 1 ) * 0.3^1 * ( 1- 0.3 ) ^9 + ( 10 0 ) * 0.3^0 * ( 1- 0.3 ) ^10   
= 0.98941
P( X > = 7 ) = 1 - P( X < 7) = 0.01059

P( X < 4) = P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 10 3 ) * 0.3^3 * ( 1- 0.3 ) ^7 + ( 10 2 ) * 0.3^2 * ( 1- 0.3 ) ^8 + ( 10 1 ) * 0.3^1 * ( 1- 0.3 ) ^9 + ( 10 0 ) * 0.3^0 * ( 1- 0.3 ) ^10
= 0.64961

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