Americans spend over $30 billion dollars annually on a variety of weight loss pr

Question

Americans spend over $30 billion dollars annually on a variety of weight loss products and services. In a study of retention rates of those using a particular weight loss program, it was found that about 16% of those who began the program dropped out in the first four weeks. Assume we have a random sample of 360 people beginning the program.

(b) What is the approximate probability that at least 300 people in the sample will still be in the program after the first four weeks? (Use the Normal approximation to the distribution. Round your answer to four decimal places.)

Americans spend over $30 billion dollars annually on a variety of weight loss products and services. In a study of retention rates of those using a particular weight loss program, it was found that about 16% of those who began the program dropped out in the first four weeks. Assume we have a random sample of 360 people beginning the program

Explanation / Answer

Mean = np = 360(0.16) = 57.6
Standard deviation = sqrt[np(1-p)]= sqrt[(360)(0.16)(1-0.16)] = 6.9558

Mean number still in the program = 360-57.6 = 302.4
Standard deviation of people still in the program = sqrt[(360)(0.84)(0.16)) =6.9558

Let x = number of people still in the program

P( at least 300 still in the program) = P( x >= 300 ):
Mean = 302.4
SD = 6.9558
Using continuity correction we use 299.5 instead of 300

= 302.4
= 6.9558
standardize x to z = (x - ) /
P(x > 299.5) = P( z > (299.5-302.4) / 6.9558)
= P(z > -0.4169) = 0.6628
(From Normal probability table)

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