Assignment on Probability and Sampling Distribution

Running Head: ASSIGNMENT ON PROBABILITY AND SAMPLING DISTRIBUTION
ASSIGNMENT ON PROBABILITY AND SAMPLING DISTRIBUTION
Name of the Student:
Name of the University:
Author Note:

ASSIGNMENT ON PROBABILITY AND SAMPLING DISTRIBUTION1
Answer 1
The proportion of basophil granulocytes is 0.6 in 100 leucocytes. Using these condition, the
fraction of basophil in 300 samples is equal to 180.Hence, there are 120 leucocytes where
basophil is absent.
According to the theory, when sample size is very large, binomial is approximated to normal
distribution with mean np and variance np(1-p), n= sample size, p=probability.
Here n=300 , p=0.4.
Thus, μ=300∗0.4=120 , σ = √300∗0.4∗0.6=8.49
Therefore,
P ( X=120 )=P ( 120−0.5< X <120+0.5 ) , 0.5 is the correction factor.
¿ P ( 119.5< X <120.5 ) =0.047(using normal table)
Hence, the probability of absence of basophil in 300 leucocytes is 0.047.
Answer 2
It is given that there are 10 cases of disease D in a year, on average. It was assumed that the
probability of a person living in that area affected with D is very low. Hence it can be said that
the probability distribution of disease infected person follows Poisson distribution with
parameter 10.
P ( X=x ) =e− λ λx
x ! , λ=10 , x=0,1,2, ...

ASSIGNMENT ON PROBABILITY AND SAMPLING DISTRIBUTION2
Probability that there are strictly less than 3 person having this disease during one year is
calculated as,
P ( X <3 ) =P ( X=0 )+ P ( X =1 ) + P ( X=2 )
¿ 0.0028Therefore, the probability that the number of person infected with disease D strictly less
than 3 is 0.0028.
Answer 3
Here it is given that the blood test results of children below 15 years follow normal distribution
with mean 22 and variance 34.In other words, if X denotes the blood test result then,
X N ¿). The density function is,
f ( x )= 1
√2 πσ e
−(x− μ)2
2 σ2
, μ=22 , σ2=34
 The proportion of children having test result higher than 25 is given as,
P ( X >25 ) =1−P ( X <25 ) =∫
−∞
25
f ( x ) dx=¿ ¿0.3035
That means 30.35% children have blood test result greater than 25.
 The fraction of students having result less than 12 is,
P ( X <12 ) =∫
−∞
12
f ( x ) dx=¿ 0.0432 ¿
Therefore, only 4.32% children have blood test result less than 12.