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Suppose that the test score of a student taking the final of a probability course is a random variable with mean 76.

Required:
a. Give an upper bound for the probability that a student's test score will exceed 86. Suppose, in addition, that the professor knows that the variance of a student’s test score is equal to 25.
(b) What can be said about the probability that a student will score between 65 and 85?
c. How many students would have to take the examination to ensure, with probability at least .9, that the class average would be within 5 of 75? Do not use the central limit theorem.

Respuesta :

Answer:

a. 0.8837

b. 0.6914

c. 10 student

Step-by-step explanation:

a. using markov's inequality; P(u >= X)<= E(u) / X

   mean u = 76

   X, test score = 86

= 76 / 86 = 0.8837

b. variance v = 25

   The test is between 65 and 85 P(65 <= X <= 85);

   P(|65 - X|<= |X - u| <= |85 - X|) = P(65 - 76 <= |X - u| <= 85 -76)

   = P(-11 <= |X - u| <= 9)

   = P(|X - u| <= 9)

using the chebyshev's inequality;

P(|X - u| <= 9) = 1 - variance / (standard deviation of individual test score)^2

   = 1 - 25/ 9^2

   = 1 - 25/81 = 1 - 0.3086 = 0.6914

c. P(|X - 76| <= 5) = 0.9

still using chebyshev's inequality;

= 1 - variance / lower limit^2 (E)

where E = 1 - 0.9 = 0.1

lower limit = 5

= 1 - 25/ 5^2 (0.1) = 10

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