In a certain community, 10% of all people above 50 years of age have diabetes. A health service in this community correctly diagnoses 95% of all persons with diabetes as having the disease, and incorrectly diagnoses 5% of all persons without diabetes as having the disease. Find the probability that a person randomly selected from among all people of age above 50 and diagnosed by the health service as having
diabetes actually has the disease.
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A random sample of 16 observations from a normal distribution yields the values
Find a 95% confidence interval for the variance of the normal distribution.
Prove that
for all m > 0, n > 0.Hence evaluate (10,7) and
.
There are three urns, urn 1 having 5 white and 3 black balls, urn II with 4 white and 4 black balls and urn III with 3 white and 5 black balls. A fair die is tossed and
i) urn I is chosen if dice shows up 1, 2 or 3,
ii) urn II is chosen if dice shows up 4 or 5,
iii) urn III is chosen if dice shows up 6.
2 balls are drawn at random from the chosen urn. What is the probability that one of the balls drawn is white and the other black?
Let X1 ,X2 ,...,Xn be a random sample from a normal population, N(.Obtain a (1-
) level confidence interval for
when
is unknown.
Let the p.d.f. of a continuous random variable X be given by
Find the maximum likelihood estimator of based on a sample X1 ,X2 ,....,Xn .
X is a random variable taking values 0 and 1 with respective probabilities q and p. A sample X1 , X2 ,...., Xn of size n is taken from the distribution.If , shows that the bias of the estimator tends to zero as n tends to infinity.
hich of the following statements are true or false? Give reasons for your answer.
i) If P(A) = 0.5, P(A È B) = 0.7 and A and B are independent events, then P(B) = .
ii) There exists a random variable X for which
.
iii) If X has the distribution N(10, 9) and if Y = 5X + 3, then Y has distribution N(53,81).
iv) The abscissa of the point of intersection of two gives of less-than type and more- than type gives the mean of the data.
v) In a problem of testing of a simple hypothesis against a simple alternative, if the probability of type 1 error is known to be 0.06, then the power of the test will be 0.94.