Suppose the standard medical treatment for a certain disease fails in 20% of all

Question

Suppose the standard medical treatment for a certain disease fails in 20% of all cases. The treatment will be given to 9 randomly selected patients with this disease.

A.) What is the distribution for the random variable X = the number of failures in a random sample of 9 patients? Bold one.

B(9, 0.2)                          B(9, 0.8)                                   N(1.8, 1.2)                              N(0, 1)

B.) On average, how many failures would we expect to see in a random sample of 9 patients? Bold one.

1 patient                        2 patients                               1.8 patients                           0.2 patients

C.) What is the probability that at most 1 of these 9 patients will result in a failure? Bold one.

0.1342                             0.3020                                      0.4364                                      0.5638                                      0.8658

D.) What is the probability that the number of failures that occur falls within one standard deviation of the expected number of failures found in (b)? Bold one.

0.6040                                     0.68                              0.7801                                  0.4782                                      Cannot tell

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
A.
B(9, 0.2)
B.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 9 * 0.2
= 1.8 patients
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 9 * 0.2 * 0.8
= 1.44
III.
standard deviation = sqrt( variance ) = sqrt(1.44
=1.2
C.
P( X < = 1) = P(X=1) + P(X=0) +   
= ( 9 1 ) * 0.2^1 * ( 1- 0.2 ) ^8 + ( 9 0 ) * 0.2^0 * ( 1- 0.2 ) ^9 +
= 0.4364
D.
NORMAL 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
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 9
standard Deviation ( sd )= 0.2
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between

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