In: Statistics and Probability
If a person is selected randomly from this group, what is the probability that he/she
Note: Final answers are highlighted in colour.
1.
a) = 100, = 5
By converting the question into Standard normal distribution,
Z=
P(X<98)= P(Z<)= P(Z<-0.4)=0.3446 (From the below Z table)
This means about 34.46% of the packets will contain less than specified on the label. This is a real concern.
b) Probability should be 0.05
That means P(X<a)= 0.05
We need to identify such a
Z value for probability of 0.05 from the Z table= -1.645
= -1.645
X= (-1.645)*(5)+ 100= 91.775
So they can change the label specifying the minimum value as 91.7
c)
I would advise the company to set its machine again in such a way that either mean increases from 100 or standard deviation reduces from 5. This way we can be more reliably promiss the minimum amount.
Also they will need to collect the number of complaints from customers and need to collect the amount of cream they received. In this way we can know the amount at which customers are getting frustrated and we can adjust the cream level to a satisfactory level.
Profit vs Customer satisfaction analysis should also be done.
2.
Total People(T)= 60
Healthy(H)= 20, Either High Blood Pressure or Cholestrol(BP U Ch)= 40
High Blood Pressure(BP)= 20, High Cholestrol(Ch)= 35.
a) People with Both High Blood Pressure and High Cholestrol(BP Ch)= (BP + Ch) - (BP U Ch)= 20+35-40=15
Therefore People with only high blood pressure()= BP - (BP Ch)= 20-15= 5
b)People that have only high level of cholesterol ()= Ch- (BP Ch)= 35- 15= 20
c)
people with high blood pressure that have high level of cholesterol too= (BP Ch) = 15
3.
a) If a person is selected randomly from this group, probability that he/she is healthy= H/T= 20/60= 1/3
b)Has high blood pressure, given that they have high level of cholesterol= P(BP|Ch)=
= (BP Ch)/ T= 15/60= 1/4
P(Ch)= Ch/T= 35/60
Therefore P(BP|Ch)= = 15/35= 0.42857
c)Has high level of cholesterol given that they do not have high level of blood pressure==
= 1-P(BP)= 1- BP/T= 1-20/60= 2/3
= = 20/60= 1/3
Therefore = = 1/2= 0.5
d) As the 2/3rd of the people eating at the cafetaria are unhealthy, I feel there is some issue with the food being served there.
If I were the medical researcher, I would urge the company to establish frequent quality checks and revoke the vendorship of some of the vendors that are producing food with bad quality.
I would ask the company to take timely customer feedbacks and assess them time to time.