2. QUESTION #1
• length (in days) of a randomly chosen human pregnancy is a normal random variable with
mean (mu, μ) = 266 and standard deviation (sigma, σ) = 16.
a) What is the probability that a randomly chosen pregnancy will last less than 246 days?
b) What is the probability that a randomly chosen pregnancy will last longer than 240
days?
3. SOLUTION:
a) Mean=266 days
S.D =16 days
P(X<246)
First we will find z-score
z = (x – μ) / σ
z=246-266/16 = -1.25
From z table P(Z<-1.25) = 0.1056 = P(X<246)
4. SOLITION
b) Mean=266 days
S.D =16 days
P(X>246)
First we will find z-score
z = (x – μ) / σ
z=240-266/16 = -1.625
From z table P(Z>-1.625) = 0.051
As it is right side we have to subtract from 1
1-0.051=0.94
P(X>246) = 0.94
5. QUESTION # 2
• The sample of 15 haematocrit values from young adult males shows the sample mean
and the sample standard deviation are 46.9 and 2.3 .
• Find P( X < 49 ).
6. SOLUTION :
Mean = 46.9
S.D = 2.3
P(X<49) .
z = (x – μ) / σ
z=(49-46.9)/2.3 = 0.91
From z table P(Z<49) = 0.818 = P(X<49)
7. QUESTION #3
• After a large number of observations of a quantitative variable (e.g. haemoglobin conc.)
have been taken from a large, random sample having mean and s.d 17 and 2.5 .
• Find P(15<X<10)
8. SOLUTION :
• Mean = 17
• S.D = 2.5 P(15<X<10)
• X = 15 & X= 10
• z = (x – μ) / σ
• Z= (15-17)/2.5 z =(10-17)/2.5
• = -0.8 =0.211 = -2.8 = 0.0025
• Now subtract smaller from larger value
• 0.211 – 0.0025 = 0.21075
• P(15<Z<10) = 0.21075 = P(15<X<10)
9. APPLICATIONS OF NORMAL DISTRIBUTION IN BMS:
• Blood pressure measurements show normal distribution . A few would have lower
b.p a few would have higher and majority`s blood pressure would be fairly close
to the average.
• Haematorict follows normal distribution.
• However, many common medical variables, such as heights, haemoglobin
concentrations, and variables from clinical chemistry have a symmetric
distribution about a single central peak, that is a Normal distribution.
10. • Note :One point that should be made is that, strictly speaking, only variables that
can take 'any' value, such as height or haemoglobin concentration, can possibly
have a Normal distribution; these are referred to as continuous variables.
Variables such as blood group or eye colour which can take only a few distinct
values, so-called discrete variables, cannot have a Normal distribution.
