Please let me know if I solved this right
The lifetime of television produced by the Hishobi Company are normally distributed with a mean of 75 months and a standard deviation of 8 months. If the manufacturer wants to have to replace only 1% of its televisions, what should its warranty be?
P (Z01<0) = 0.5 – 0.01 =.4900 2.33 × 8 =18.64 75 - 18.64 = 56.36 X=56.36
The Z score should be -2.33 making the score 56.36
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score related to that proportion (.01).
Z = 2.33 (approx. for .01)
Z = (score-mean)/SD
2.33 = (score-75)/8
Multiply both sides by 8
18.64 = score - 75
ADD 75 to both sides.
93.64 = score
Well, if we want to find the warranty period for the televisions, we need to find the value of X, which represents the number of months we want the warranty to cover. It seems like you've already done all the calculations, but let's break it down in a fun way!
So, we know that the mean lifetime for the televisions is 75 months, and the standard deviation is 8 months. We want to find the warranty period that covers only 1% of the televisions, meaning we want to find the X value such that only 1% of the televisions will fail before that time.
To find this X value, we can use the Z-score formula: Z = (X - Mu) / Sigma. Here, Mu is the mean (75 months) and Sigma is the standard deviation (8 months). We want to find the Z-score corresponding to a cumulative probability of 0.01.
But fear not! You don't have to calculate the Z-score by yourself – you can use a standard normal distribution table or a statistical calculator. In this case, the Z-score that corresponds to a cumulative probability of 0.01 is approximately -2.33.
Now, we use the formula Z = (X - Mu) / Sigma and plug in the values we know: -2.33 = (X - 75) / 8. Rearranging the equation, we have X - 75 = -2.33 * 8.
Simplifying, we find that X - 75 = -18.64, and by adding 75 to both sides, we get X = 56.36.
So, the warranty period for the televisions should be approximately 56.36 months.
Now, please keep in mind that this calculation assumes a normal distribution, and there may be additional factors to consider in real-world scenarios. But for the purposes of this exercise, this should give you a reasonable estimate.
To calculate the warranty period, we need to find the value on the distribution curve that corresponds to the 1% point on the left side, denoted as Z0.01.
The formula to calculate Z0.01 is:
Z0.01 = mean - (Z * standard deviation)
Since the question does not specify a one-tailed or two-tailed test, we assume a one-tailed test.
Now, let's calculate Z0.01:
Z0.01 = 0.5 - 0.01 = 0.49
To find the actual value for the warranty, we multiply Z0.01 by the standard deviation and subtract the result from the mean:
X = mean - (Z * standard deviation)
X = 75 - (0.49 * 8)
X = 75 - 3.92
X ≈ 71.08
Therefore, the warranty period should be approximately 71.08 months for the Hishobi televisions to replace only 1% of them.
To determine the warranty period for the television, we need to find the value of X that corresponds to the 1% (or 0.01) percentile in a standard normal distribution.
The 1% percentile can also be written as Z = -2.33 (since the standard normal distribution has a mean of 0 and a standard deviation of 1, and the 1% percentile corresponds to a Z-score of -2.33).
Next, we multiply the Z-score (-2.33) by the standard deviation (8) to get -2.33 * 8 = -18.64. This tells us that 18.64 months below the mean lies the 1% percentile value.
To find the warranty value, we subtract 18.64 from the mean (75) to get 75 - 18.64 = 56.36. Therefore, the warranty should be set to approximately 56.36 months.