The drug Zyban is meant to create the urge to quit smoking. In clinical trials, 35% of the study's participants experienced insomnia when taking 300mg of Zyban per day. a) In a random sample of 25 users of Zyban, find the probability that at least 5 will experience insomnia. b) In a random sample of 25 users of Zyban, find the probability that at most 12 will experience insomnia. 3. Find the mean and standard deviation of x = number of cars found to have unsafe brakes among the 500 cars stopped at a roadblock for inspection. Assume that 5% of all cars have one or more unsafe brakes. a) Mean = b) Standard deviation =

Step-by-step explanation:

Hello!

1) Trial of drug Zyban

The variable of interest is

X: number of participants that experienced insomnia after taking a 300 mg Zyban per day.

The proportion obtained during the trial is p= 0.35

This variable has a binomial distribution X~Bi(n;p), where the "success" of the binomial experiment is that the participant experienced insomnia.

a)

You have to calculate the probability of at least 5 participants out of 25 to experience insomnia. Symbolically:

P(X ≥ 5)

Using the binomial tables with cumulative probabilities for a distribution X~Bi(25;0.35) you have to look for the probability accumulated until 5 successes, then you subtract this probability from 1 to obtain the probability above 5:

P(X ≤ 5) = 0.08

P(X ≥ 5)= 1 - P(X ≤ 5) = 1 - 0.08= 0.92

b)

Now you have to calculate the probability of at most 12 participants to experience insomnia for X~Bi(25;0.35), you can obtain this value directly from the table:

P(X ≤ 12)= 0.939= 0.94

2) Cars brake effectiveness

The variable of interest is

X: Number of cars found to have unsafe brakes among 500 cars stopped at a roadblock for inspection.

The proportion of cars that have one or more unsafe brakes is p=0.05

n= 500 cars.

Checking the binomial criteria, the number of trials is fixed, n=500 cars, only two possible outcomes "Success": the brakes are unsafe, and "failure": the brakes are safe. Each car is independent of the others, and the probability of success is the same for all the trials.

The mean of the binomial distribution is calculated as:

E(X)= n*p= 500*0.05= 25

Variance and standard deviation:

V(X)= n*p*(1-p)= 500*0.05*0.95= 23.75

The standard deviation is the square root of the variance:

√V(X)= √23.75= 4.87

I hope this helps!

i need a picture

step-by-step explanation: