Find probability that a randomly selected TV will have replacement time less than 6 yrs.
mean is 8.2 yrs
standard deviation is 1.1
then provide warranty of 1% will be replaced what is the time length of the warranty
To find the probability that a randomly selected TV will have a replacement time less than 6 years, you can use the concept of the standard normal distribution.
First, you need to standardize the value of 6 years to find its z-score. The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- x is the value you want to standardize (in this case, 6 years),
- μ (mu) is the mean (average) replacement time (8.2 years), and
- σ (sigma) is the standard deviation (1.1 years).
Plugging in the given values:
z = (6 - 8.2) / 1.1
z = -2 / 1.1
z ≈ -1.82
Once you have the z-score, you can use a standard normal distribution table or a calculator/statistical software to find the corresponding probability. The z-score represents the number of standard deviations a value is from the mean, so you want to find the probability of an event happening within that range.
Looking up the z-score of -1.82 in a standard normal distribution table, you will find that the probability (P) is approximately 0.0344.
Therefore, the probability that a randomly selected TV will have a replacement time less than 6 years is approximately 0.0344, or 3.44%.
Now, for the second part of your question, if a warranty replacement occurs in only 1% of cases, this implies that the warranty covers the TVs until the replacement time that corresponds to the z-score for the upper 1% of the standard normal distribution.
To find this z-score, you look for the area under the curve to the left of it that corresponds to 99% (100% - 1%). In other words, you need to find the z-score that leaves a probability of 0.99 to the left of it.
Using a standard normal distribution table, you can find this value. The z-score that corresponds to a 99% probability is approximately 2.33.
To get the time length of the warranty, you need to convert this z-score back into years using the mean and standard deviation.
warranty time = (z * σ) + μ
warranty time = (2.33 * 1.1) + 8.2
warranty time ≈ 10.563 years
Therefore, the time length of the warranty would be approximately 10.563 years.