Page 220 - DMTH404_STATISTICS

Statistics

Notes Solution.

Let the mean and standard deviation of the given distribution be m and s respectively.

-
æ 4500 m ö
It is given that (X < 4500 ) = 0.0228 or P z < ÷ = 0.0228
P
ç
è  ø
On locating the value of z corresponding to an area 0.4772 (0.5000 - 0.0228), we can write
4500 m
-
-
= - 2 or 4500 m = - 2 .... (1)

Similarly, it is also given that
-
æ 7500 m ö
P (X > 7500 ) = 0.1587 or P z >  ÷ ø = 0.1587
ç
è
Locating the value of z corresponding to an area 0.3413 (0.5000 - 0.1587), we can write
-
7500 m
-
=
= 1 or 7500 m  .... (2)

Solving (1) and (2) simultaneously, we get
m = Rs 6,500 and s = Rs 1,000.

Example 33: Marks in an examination are approximately normally distributed with
mean 75 and standard deviation 5. If the top 5% of the students get grade A and the bottom 25%
get grade F, what mark is the lowest A and what mark is the highest F?
Solution.
Let A be the lowest mark in grade A and F be the highest mark in grade F. From the given
information, we can write

æ A - 75ö
P (X  A ) = 0.05 or P z  ÷ = 0.05
ç
è 5 ø
On locating the value of z corresponding to an area 0.4500 (0.5000 - 0.0500), we can write
A - 75
= 1.645 Þ A = 83.225
5
Further, it is given that

æ F - 75ö
P (X £ ) F = 0.25 or P z £ ÷ = 0.25
ç
è 5 ø
On locating the value of z corresponding to an area 0.2500 (0.5000 - 0.2500), we can write
F - 75
= - 0.675 Þ F = 71.625
5

Example 34: The mean inside diameter of a sample of 200 washers produced by a machine
is 5.02 mm and the standard deviation is 0.05 mm. The purpose for which these washers are
intended allows a maximum tolerance in the diameter of 4.96 to 5.08 mm, otherwise the washers
are considered as defective. Determine the percentage of defective washers produced by the
machine on the assumption that diameters are normally distributed.

212 LOVELY PROFESSIONAL UNIVERSITY