Question2: (Marks:7+3= 10) a) A random variable X has the following probabilityd
ID: 2913450 • Letter: Q
Question
Question2: (Marks:7+3= 10)
a)
A random variable X has the following probabilitydistribution:
x
4 5 6
P(x)
0.3 0.5 0.2
Find the mean and standard error of the mean for a randomsample of size 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.
Question3: (Marks:5x2= 10)
a)
In a random sample of 500 people eating lunch at a hospitalcafeteria on various Fridays, it was found that preferred seafood.Find 95% confidence interval for the actual proportion of peoplewho eat seafood on Fridays at this cafeteria.
b)
The mean and standard deviation for the quality grade-pointaverages of a random sample are calculated to be 2.6 and 0.3. Howlarge sample is required if we want to be 95% confident that ourestimate will not be in error by more than 0.05.
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) = 0.00763 2)in the second problem, number of people preferring seafoodis not given if X is the number of people preferring sea food , then theproportion of people preferring sea food , p = X/n then the 95% confidence interval is p ±Z/2(pq/n)Related Questions
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