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A study of a hypertension drug’s effectiveness is being conducted. It is considered that the probability that 0 patients of 4 will be brought under control is 0.006, 1 out of 4 brought under control is 0.078, 2 of 4 is 0.261, 3 of 4 is 0.415, and all 4 is 0.240.
The probability mass function table is given in the excel sheet. Let X be the random variable representing the number of patients whose hypertension is brought under control. Find the following (5 pts each):
(a) Set up the distribution table (here and in Excel)
(b) Find E(X)
(c) Find E(X2)
(d) Find V (X)
(e)Find the standard deviation of X.
(a) | X | 0 | 1 | 2 | 3 | 4 | ||||
P(X) | 0.006 | 0.078 | 0.261 | 0.415 | 0.24 | |||||
(b) | X*P(X) | 0 | 0.078 | 0.522 | 1.245 | 0.96 | E(X)=ΣX*P(X) | |||
E(X)= | 2.805 | |||||||||
( c ) | X^2 | 0 | 1 | 4 | 9 | 16 | E(X^2)=ΣX^2*P(X) | |||
P(X) | 0.006 | 0.078 | 0.261 | 0.415 | 0.24 | |||||
X^2*P(X) | 0 | 0.078 | 1.044 | 3.735 | 3.84 | |||||
E(X^2)= | 8.697 | |||||||||
(d) | V(X)= | E(X^2)-(E(X))^2) | ||||||||
V(X)= | 0.828975 | |||||||||
( e ) | S.D(X)= | sqrt(V(X)) | ||||||||
S.D(X)= | sqrt(0.828975) | |||||||||
0.910480642 | ||||||||||
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