SOLUTION: B1. The average lifetime of a light bulb is 3,000 hours with a standard deviation of 696 hours. A simple random sample of 36 bulbs is taken. a. What is the probability that the

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Question 1135627: B1. The average lifetime of a light bulb is 3,000 hours with a standard deviation of 696 hours. A simple random sample of 36 bulbs is taken.
a. What is the probability that the average life in the sample will be between 2,670.56 and 2,809.76 hours?
b. What is the probability that the average life in the sample will be greater than 3,219.24 hours?

B2. The Minneapolis Star Tribune (August 12, 2008) reported that 73% of Americans say that Starbucks coffee is overpriced. The source of this information was a national telephone survey of 100 American adults conducted by Rasmussen Reports.
(a) What is the probability that the proportion of people that will say the coffee is overpriced is between .68 and .78?
(b) What is the probability that more than 85% of people will say the coffee is overpriced?

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
question B1:

The average lifetime of a light bulb is 3,000 hours with a standard deviation of 696 hours. A simple random sample of 36 bulbs is taken.

average life is 3000 hours with standard deviation of 696 hours.

sample size is 36

standard error = s = 696 / sqrt(36) = 116

question B1 part a:

What is the probability that the average life in the sample will be between 2,670.56 and 2,809.76 hours?

using the normal distribution calculator at http://davidmlane.com/hyperstat/z_table.html, you can get the answer directly as shown below.

$$$

the answer is that the probability is equal to .0482.

you need to use the standard error and not the standard deviation.

question B1 part 2:

What is the probability that the average life in the sample will be greater than 3,219.24 hours?

using the normal distribution calculator at http://davidmlane.com/hyperstat/z_table.html, you can get the answer directly as shown below.

$$$

the answer is that the probability is equal to .0294.

using the referenced calculator is the easy way to do it.

if you need to use the normal distribution tables to get the answer manually, let me know and i'll guide you through that, but if you just need the answer in the easiest way possible to get it, the referenced calculator does the trick very nicely.

question B2:

The Minneapolis Star Tribune (August 12, 2008) reported that 73% of Americans say that Starbucks coffee is overpriced. The source of this information was a national telephone survey of 100 American adults conducted by Rasmussen Reports.

p = .73
q = 1 - p = .27
n = 100

m = p = .73
s = sqrt(p * q / n) = sqrt(.73 * .27 / 100) = .0443959 rounded to 7 decimal places.

question B2 part a:

What is the probability that the proportion of people that will say the coffee is overpriced is between .68 and .78?

using the normal distribution calculator at http://davidmlane.com/hyperstat/z_table.html, you can get the answer directly as shown below.

$$$

the probability that between .68 and .78 proportion of the population will say that the coffee is over-priced is equal to .7399.


question B2 part b:

What is the probability that more than 85% of people will say the coffee is overpriced?

using the normal distribution calculator at http://davidmlane.com/hyperstat/z_table.html, you can get the answer directly as shown below.

$$$

the probability that more than 85% of the people will say the coffee is over-priced is equal to .0035.

with a proportion type problem, the mean is equal to p and the standard error is equal to sqrt(p * q / n).

in this problem, p is the mean proportion of .73 and the standard error is equal to sqrt(.73 * .27 / 100)







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