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.

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

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

the highest percentage for deaths in construction from highest to lowest is as follows: falls, struck by objects, electrocution, and caught-in/between. these four are collectively known as fatal four.

in the case of falls, the type that is responsible for the overall highest death in the category is falls from roofs.