Hi! I've been stuck on this question for a while. I know that I have to do some sort of 1-1/k^2 for empirical rule, correct? I can't figure out how to calculate any of this and what it means of within the standard deviation. I would greatly appreciate any help on this - thank you very much!
The mean life of a brand of auto batteries is 44 months with a standard deviation of 3 months. Assume that the lives of all auto bateries of this brand have a bell-shaped distribution.
1. using empirical rule, find the percentage of auto batteries of this brand that have a life of
a. 41 to 47 months
b. 35 to 53 months
2. using empirical rule, find the interval that contains the mean life of the batteries by 95% of the batteries.
3. One customer says that he used the battery for 52 months. Find the z-score for the 52 month.
I don�ft know about the 1-1/k^2 empirical rule. However, you can answer all of the questions by using Z scores.
Z = (X - ƒÊ)/SD
where X = raw score, ƒÊ = mean and SD = standard deviation.
Figure the Z scores for each of the raw scores and look up the percentages cut off in the normal distribution. This can be found in a table in the back of your statistics textbook.
I hope this helps. Thanks for asking.
To solve these questions, we will calculate the z-scores and then use the z-table to find the corresponding percentages.
1. Percentage of auto batteries with a life of 41 to 47 months:
We will calculate the z-scores for 41 and 47.
For 41 months:
Z = (X - ƒÊ) / SD
Z = (41 - 44) / 3
Z = -1
For 47 months:
Z = (X - ƒÊ) / SD
Z = (47 - 44) / 3
Z = 1
Using the z-table, we can find the percentage of batteries within the range of -1 to 1. This represents approximately 68% of the batteries.
2. Interval containing the mean life of batteries by 95%:
We need to find the z-score corresponding to the percentage of 95%. Since 95% of the data is within two standard deviations of the mean, we need to find the z-score that corresponds to 0.95.
Using the z-table, we find that the z-score for a cumulative area of 0.95 is approximately 1.96.
Now, we can calculate the interval around the mean life of batteries:
Lower limit = ƒÊ - (SD * z-score)
Upper limit = ƒÊ + (SD * z-score)
Lower limit = 44 - (3 * 1.96)
Lower limit = 44 - 5.88
Lower limit ≈ 38.12
Upper limit = 44 + (3 * 1.96)
Upper limit = 44 + 5.88
Upper limit ≈ 49.88
Therefore, the interval containing the mean life of batteries by 95% of the batteries is approximately 38.12 to 49.88 months.
3. Z-score for a battery life of 52 months:
We can calculate the z-score using the formula mentioned earlier.
Z = (X - ƒÊ) / SD
Z = (52 - 44) / 3
Z = 2.67
The z-score for a battery life of 52 months is approximately 2.67.
I hope this helps! Let me know if you have any further questions.
1. To find the percentage of auto batteries that have a life of 41 to 47 months, you need to calculate the z-scores for these values using the formula:
Z = (X - μ)/σ
where X is the raw score (41 or 47), μ is the mean (44), and σ is the standard deviation (3).
For 41 months:
Z = (41 - 44)/3 = -1
For 47 months:
Z = (47 - 44)/3 = 1
Now, you can look up the percentage of data that falls within these z-scores in a standard normal distribution table. The empirical rule states that approximately 68% of the data falls within 1 standard deviation of the mean, so the percentage of auto batteries with a life of 41 to 47 months would also be around 68%.
For the interval of 35 to 53 months, you can follow the same steps:
For 35 months:
Z = (35 - 44)/3 = -3
For 53 months:
Z = (53 - 44)/3 = 3
Since the empirical rule states that approximately 95% of the data falls within 2 standard deviations of the mean, the percentage of auto batteries with a life of 35 to 53 months would be around 95%.
2. To find the interval that contains the mean life of the batteries by 95% of the batteries, you can use the formula:
Z = ± (1.96 × σ)
Since 1.96 falls within the 95% confidence interval, you can calculate the interval as:
Lower limit = μ - (1.96 × σ)
Upper limit = μ + (1.96 × σ)
For the given example:
Lower limit = 44 - (1.96 × 3) = 37.12
Upper limit = 44 + (1.96 × 3) = 50.88
Therefore, the interval that contains the mean life of the batteries by 95% of the batteries is approximately 37.12 to 50.88 months.
3. To find the z-score for a battery life of 52 months, you can use the formula:
Z = (X - μ)/σ
where X is the raw score (52), μ is the mean (44), and σ is the standard deviation (3).
Z = (52 - 44)/3 = 2.67
This means that the battery life of 52 months is approximately 2.67 standard deviations above the mean.