Suppose that the lifetimes of old-fashioned TV tubes are normally distributed with a standard deviation of 1.2 years. Suppose also that exactly 40% of the TV tubes die before 4 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.years
Question
Suppose that the lifetimes of old-fashioned TV tubes are normally distributed with a standard deviation of 1.2 years. Suppose also that exactly 40% of the TV tubes die before 4 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.years
Solution
To solve this problem, we need to use the properties of the normal distribution.
-
First, we know that 40% of the TV tubes die before 4 years. This means that the z-score corresponding to 40% (or 0.40) is the number of standard deviations that 4 years is away from the mean.
-
We can look up 0.40 in a standard normal distribution table or use a z-score calculator to find the corresponding z-score. The z-score for 0.40 is approximately -0.2533.
-
The z-score formula is Z = (X - μ) / σ, where Z is the z-score, X is the value from the dataset, μ is the mean, and σ is the standard deviation. We can plug in the values we know: -0.2533 = (4 - μ) / 1.2.
-
Solving for μ, we get μ = 4 - (-0.2533 * 1.2) = 4 + 0.30396 = 4.304.
So, the mean lifetime of TV tubes is approximately 4.3 years.
Similar Questions
Suppose a manufacturer wants to estimate the averagelifespan of a new product they have developed. Theyrandomly select 15 products and test them until they fail.It is known that the average lifespan is 120 hours, with astandard deviation of 10 hours. Assume that the lifespanof the products follows a normal distribution. Themanufacturer wants to know the probability that theaverage lifespan of the product is greater than 130 hours.
The manufacturer of television tubes knows from past experience that the average life of a tubeis 2000 hrs. with a s.d. of 200 hrs. A sample of 100 tubes has an average life of 1950 hrs. Test atthe 0.01 level of significance to see if this sample came from a normal population of mean 2000hrs
A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 8989 months with a standard deviation of 55 months.If the claim is true, what is the probability that the mean monitor life would be greater than 87.887.8 months in a sample of 123123 monitors? Round your answer to four decimal places.
A factory produces a type of light tube, Its lifespan can be expressed this way: X~(u, 200^2). From experience, it's known that the average lifespan u is 1500 hours. Now, after adopting a new technique, 25 light tubes are randomly selected from those produced and the average lifespan is found to be 1675 hours. With the new technique adopted, what is the value of the test statistic?
Suppose that the lifetimes of tires of a certain brand are normally distributed with a mean of 74,000 miles and a standard deviation of σ miles. These tires come with a 55,000-mile warranty. The manufacturer of the tires can adjust σ during the production process, but the adjustment of σ is quite costly. The manufacturer wants to set σ once and for all so that only 5% of the tires will fail before warranty expires. Find the standard deviation to be set. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.miles
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.