The study at www.state.sd.us/DOH/Nutrition/TV.pdf showed men watched TV with a m

Question

The study at www.state.sd.us/DOH/Nutrition/TV.pdf showed men watched TV with a mean of 26 hours per week and standard deviation 4.6 hours. Assume that the distribution of hours watched follows the normal distribution. We want to know the probability that a random man watches less than 19 hours of TV per week. a) What is the z-score? Use 2 decimal places b) What is the probability a random man watches less than 19 hours of TV per week? Usc 4 decimal placcs Now suppose you are a woman who wishes to determine your chances of finding a man who is rarely watches any TV. Determine the probability that you will randomly select a man who watches less than 12 hours of TV per week. Use 4 decimal places

Explanation / Answer

a) z score = (19 - 26)/4.6 = -1.52

b) P(X < 19)

= P(z < -1.52)

= 0.0640

c) P(X< 12)

= P(z < (12 - 26)/4.6)

= P(z < -3.04)

= 0.0017

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