Question

According to a 2009 Reader's Digest article, people throw away approximately 20% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 200 grocery shoppers to investigate their behavior. What is the probability that the sample proportion is between 0.05 and 0.13? Note: You should carefully round any intermediate values you calculate to 4 decimal places to match wamap's approach and calculations. Answer = (Enter your answer as a number accurate to 4 decimal places.)

The physical plant at the main campus of a large state university receives daily requests to replace florescent light bulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 56 and a standard deviation of 7. Using the empirical (68-95-99.7) rule, what is the approximate percentage of light bulb replacement requests numbering between 35 and 56? Do not enter the percent symbol. Enter your answer has a number between 1 and 100 (not between 0 and 1). For example, for 23.6%, you would enter 23.6 into the box. Answer = %

Answer 1

1) p = 0.2

   n = 200

= p = 0.2

= sqrt(p(1 - p)/n)

      = sqrt(0.2 * 0.8/200)

      = 0.0283

P(0.05 < < 0.13)

= P((0.05 - )/ < ( - )/ < (0.13 - )/)

= P((0.05 - 0.2)/0.0283 < Z < (0.13 - 0.2)/0.0283)

= P(-5.3 < Z < -2.47)

= P(Z < -2.47) - P(Z < -5.3)

= 0.0068 - 0

= 0.0068

2) 56 - 3 * 7 = 35

   According to the empirical rule about 99.7% of the data fall within 3 standard deviation from the mean.

35 is 3 standard deviation below the mean.

S0 49.85% of light bulb replacement requests numbering between 35 and 56.