Posts tagged with normal approximation to binomial

Qn18 Find the indicated probability using the standard normal distribution
P(-1.16 < z < 1.16)
Qn19. Assume the random variable x is normally distributed with mean mu = 50 and standard deviation sigma = 7. Find the indicated probability.
P(x > 43)
Qn20. Use the normal distribution of SAT critical reading scores for which the mean is 514 and the standard deviation is 121. Assume the variable x is normally distribted.
a. What percentage of the SAT verbal scores are less than 550?
b. If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 525?

MyMathLab Homework 5.5 Answers
A drug tester claims that a drug cures a rare skin disease 78% of the time. The claim is checked by testing the drug on 100 patients. If at least 73 patients are cured, the claim will be accepted.

Find the probability that the claim will be rejected assuming that the manufacturer's claim is true. Use the normal distribution to approximate the binomial distribution if possible.

Solving the question

The first step for solving this question is to check that the conditions for normal to binomial approximation are met.,
np >= 5, and nq >= 5

In this case the conditions are met because np; 1000.78 = 78 , & nq = 0.22100 = 22
q = 1-p; = 1-0.78 = 0.22
To use the normal approximation , we need to calculate the mean and standard deviation based on the data given.
np = 1000.78 = 78, and standard deviation sqrt(npq) = sqrt(0.780.22*100) = 4.142

We are interested in the probability that 1-p(X < 73)
We'll use continuity correction to improve the accuracy of our calculations.
Therefore, we'll find p(X < 72.5)

The probability will be rejected if the value is less than 73.
P(X < 73) = (72.5 - 78)/4.142 = -1.33
P(X < -1.33) = 0.0921.,

Note that you can also use Excel to answer this question,
the function; =NORM.DIST(72.5,78,4.142,TRUE) is very helpful when dealing with such kinds of questions.

You are free to contact MyMathLab Homework Help in case you need help in answering similar questions.