00:01
In question 5, it is given that 25 % of the patient who has undergoes a medical procedure has a side effects and it is given that n equal to 60 patients are included in a clinical trial.
00:20
Now, in part a, we have to test whether the sample proportion can be approximated to the normal distribution or not.
00:30
Now using the central limit theorem as the theorem states that as the sample size n increases, then the sample proportion p cap tends to follow the normal distribution with the mean mu of p -cap equal to the population proportion p which is given as the 25 percentages 0 .25 and with the standard deviation, sigma of p -cap equal to root of p into 1 minus p divided by n substituting would be give root of 0 .25 into 1 minus 0 .25 divided by 60.
01:14
Now simplifying this gives the sigma p -cap value of 0 .056.
01:21
Thus it can be appropriate that the sample distribution of the sample proportion tends to follow the normal distribution thus the correct option is yes in part b we have to find the probability that less than 20 people have the side effects first we have to find the sample proportion for this 20 people and it is calculated as the given 20 people divided by the total number of people in the clinical trial which is 60.
01:59
Now simplifying this gives the sample proportion value of 0 .33 -33.
02:06
Now we have to find the probability that the sample proportion p -cap is less than 0 .33 -douple -3.
02:16
And this can be graphically represented as we have to find the area which is less than 0 .3323, that is the area to the left of 0 .333.
02:32
And this can be calculated as probability that to standardize the value, we have to find p -cap minus mu of p -cap divided by sigma of p -cap, which is less than 0 .33 -2 -3 -mine, minus the mu -cap value is 0 .25.
02:53
Divided by the sigma of p -cap value is found as 0 .056.
02:59
Now simplifying this gives the value of probability that z is less than 1 .49.
03:07
Now to find this probability, the standard normal distribution table is given below...