a) A random variable X has the following probabilitydistribution: x 4 5 6 P(x) 0
ID: 2913451 • Letter: A
Question
a)
A random variable X has the following probabilitydistribution:
x
4 5 6
P(x)
0.3 0.5 0.2
b)
It is known that 3% of the persons living in Gujranwala city areknown to have a certain disease. Find the mean and standard errorof sampling distribution of proportion of diseased persons in arandom sample of 500 persons.
x
4 5 6
P(x)
0.3 0.5 0.2
Explanation / Answer
given tha data, x 4 5 6 p(x) 0.3 0.5 0.2 mean = 4*0.3+5*0.5+6*0.2 =1.2 +2.5 + 1.2 = 4.9 standard deviation = sqrt( x^2*p(x) - (x*p(x))^2) =sqrt([16*0.3+25*0.5+36*0.2 ] - [4.9]^2) =sqrt( 24.5 - 24.01) =sqrt( 0.49) =0.7 b)the probability of persons having disease, p =3% =0.03 number of persons, n = 500 mean of sampling distribution of proportion, np= 0.03*500 = 15 standard error ofproportion, SE(p) = sqrt( pq/n) =sqrt( 0.03*0.97 / 500) =sqrt( 0.0000582) = 0.00763Related Questions
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