Page 222 - DMTH404_STATISTICS

Statistics

Notes Solution.

Let X denote the percentage of marks and its mean and S.D. be m and  respectively. From the
given table, we can write
P(X < 40) = 0.43 and P(X  75) = 0.15, which can also be written as

-
-
æ 40 m ö æ 75 m ö
P z < ÷ = 0.43 and P z  ÷ = 0.15
ç
ç
è  ø è  ø
The above equations respectively imply that
-
40 m
= - 0.175 or 40 m = - 0.175 .... (1)
-

75 m
-
-
and = 1.04 or 75 m = 1.04 .... (2)

Solving the above equations simultaneously, we get m = 45.04 and  = 28.81.
Let X be the percentage of marks required to pass the examination.
1
æ X - 45.04ö
1
P
Then we have (X < X 1 ) 0.3 or P z= ç è < 28.81 ÷ ø = 0.3
X - 45.04
 1 = - 0.525 Þ X = 29.91 or 30% (approx.)
1
28.81
Example 37: In a certain book, the frequency distribution of the number of words per
page may be taken as approximately normal with mean 800 and standard deviation 50. If three
pages are chosen at random, what is the probability that none of them has between 830 and 845
words each?
Solution.
Let X be a normal variate which denotes the number of words per page. It is given that
X ~ N(800, 50).
The probability that a page, select at random, does not have number of words between 830 and
845, is given by

-
-
æ 830 800 845 800ö
1 P (830 < X < 845 ) 1 P= - ç < z < ÷
-
è 50 50 ø
=1- P 0.6 < z <0.9f =1- P 0 < z <0.9f+P 0 < z <0.6f
a
a
a
= 1- 0.3159 + 0.2257 = 0.9098 » 0.91
Thus, the probability that none of the three pages, selected at random, have number of words
3
lying between 830 and 845 = (0.91) = 0.7536.
Example 38: At a petrol station, the mean quantity of petrol sold to a vehicle is 20 litres
per day with a standard deviation of 10 litres. If on a particular day, 100 vehicles took 25 or more
litres of petrol, estimate the total number of vehicles who took petrol from the station on that
day. Assume that the quantity of petrol taken from the station by a vehicle is a normal variate.

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