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2.7.3.Different forms of area tables
Unit 2 - Probability distributions
2.7.3.Different forms of area tables
Area under the normal curve is available in tables in different forms. For instance:
(i) In Fisher and Yates (1963) the area of normal curve is tabulated from Z to
Fig. a: Area from Z to
(ii) In Spiegel (1981) area of the normal curve is tabulated from 0 to any positive value of Z.
Fig.b: Area from 0 to Z
(iii) In Woolf (1968) area of the normal curve is tabulated from - to Z.Fig.c: Area from - to Z
Note:
Before referring to these tables, it is therefore necessary to know the manner in which areas are presented.
In the present manual, area tables as presented in Spiegel (1981) are referred to.
Example 8
The mean length of a one year old brood of catla is 30 cm and standard deviation 2 cm. A fish is caught at random, find the probability that its length is,
(i) (a) Between 30 and 32cm
(b) Between 28 and 33 cm
(ii) Suppose it was decided to transfer all those having length greater than 31cm, what percent of fish is required to be transferred? Assume lengths are normally distributed.
Answer
(i) (a) Compute the corresponding standard normal variate Z, for X1=30 and X2 =32
They are,
P (0≤Z≤1) = Area between (Z=0 and Z=1) =0.3413
The area is obtained by referring to the area table of normal distribution.
(b) Length between 28 and 33 cm i.e. X1= 28, X2= 33, corresponding standard normal variates are,
The probability that the length of the fish caught is between 28 and 33 cm in terms of Z will be
P (-1≤Z≤1.5) = Area between Z = -1 and Z = 1.5
= (Area between Z = -1 and Z=0) + (Area between Z = 0 and Z = 1.5)
= (Area between Z = 0 and Z = 1) + (Area between Z = 0 and Z = 1.5)
= 0.3413 + 0.4332
= 0.7745
(ii) Here X1 = 31 cm hence,
= P (Z>0.5)
= (Area to the right of Z=0) - (Area between Z=0 and 0.5)
= 0.5 – 0.1915
= 0.3085
Therefore, 30.85% of the fishes require to be transferred.
Last modified: Monday, 12 September 2011, 8:52 AM