Can we do 6 and 7 please? Problem(6) (3 points) The probability that a patient r

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Can we do 6 and 7 please?

Problem(6) (3 points) The probability that a patient recovers from a rare blood disease is 30%. If 14 people are known to have contracted this disease, what is the probability that (a) at least 10 survive, (b) from 4 to 8 survive, and (c) exactly 7 survive ? Problem(7) (3 points) Assume that the mean age of the students at UHD is 28.3 years, and the standard deviation is 12 years. A random sample of 100 students is drawn. What is the probability that the average age of these students is greater than 30 years? ntol Tot Y Normalus 25) andZNormal(0, 1), use

Explanation / Answer

QUESTION 6.
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 = 14 * 0.3
= 4.2
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 14 * 0.3 * 0.7
= 2.94
III.
standard deviation = sqrt( variance ) = sqrt(2.94)
=1.714643
a.
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) +  
= ( 14 9 ) * 0.3^9 * ( 1- 0.3 ) ^5 + ( 14 8 ) * 0.3^8 * ( 1- 0.3 ) ^6 + ( 14 7 ) * 0.3^7 * ( 1- 0.3 ) ^7 + ( 14 6 ) * 0.3^6 * ( 1- 0.3 ) ^8 + ( 14 5 ) * 0.3^5 * ( 1- 0.3 ) ^9 + ( 14 4 ) * 0.3^4 * ( 1- 0.3 ) ^10 + ( 14 3 ) * 0.3^3 * ( 1- 0.3 ) ^11 + ( 14 2 ) * 0.3^2 * ( 1- 0.3 ) ^12 + ( 14 1 ) * 0.3^1 * ( 1- 0.3 ) ^13 + ( 14 0 ) * 0.3^0 * ( 1- 0.3 ) ^14
= 0.998334
P( X > = 10 ) = 1 - P( X < 10) = 0.001666
b.
P(X=4)+ P(X=5) +P(X=6)+P(X=7)+P(X=8)
=0.229034+0.196315+0.126202+0.061813+0.02318
=0.636544
c.
P( X = 7 ) = ( 14 7 ) * ( 0.3^7) * ( 1 - 0.3 )^7
= 0.061813

QUESTION 7.
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
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 28.3
standard Deviation ( sd )= 12/ Sqrt ( 100 ) =1.2
sample size (n) = 100
P(X > 30) = (30-28.3)/12/ Sqrt ( 100 )
= 1.7/1.2= 1.4167
= P ( Z >1.4167) From Standard Normal Table
= 0.0783

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