Example:
A manufacturer of TV sets wants to advertise that if a TV manufactured by the company lasts for less than a certain number of years, the company will refund the full amount paid for the set. The company wants to pick the number of years for the
advertisement in such a way that it will not have to give refunds on more than 4% of the sets. If the life of a TV set is normally distributed with a mean life of 8.5 years and a standard deviation of 1.8 years, provide the number of years for the
company’s advertisement.
Answer:
x = 8.5 + (-1.75)(1.8) = 5.35 years; so that P(X < x) = 0.0400
2. The cumulative area increases as the z-scores increase.
3. The cumulative area for z=0 is .5000.
4. The cumulative area is close to one for z-scores close to z-.349.
A manufacturer of TV sets wants to advertise that if a TV manufactured by the company lasts for less than a certain number of years, the company will refund the full amount paid for the set. The company wants to pick the number of years for the
advertisement in such a way that it will not have to give refunds on more than 4% of the sets. If the life of a TV set is normally distributed with a mean life of 8.5 years and a standard deviation of 1.8 years, provide the number of years for the
company’s advertisement.
Answer:
x = 8.5 + (-1.75)(1.8) = 5.35 years; so that P(X < x) = 0.0400
- The standard normal distribution is a normal distribution with a mean of zero and a standard deviation of one.
- Properties of the Standard Normal Distribution
2. The cumulative area increases as the z-scores increase.
3. The cumulative area for z=0 is .5000.
4. The cumulative area is close to one for z-scores close to z-.349.