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Question 1 of 5
1. Question
`20 000` light bulbs are tested to see how long they last. The collected data is normally distributed. If `95%` of the light bulbs lasted within `29 800` and `31 100` hours, what is the standard deviation? The mean is `30,000`.Hint
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Given Values
Mean`= 30 000`This means that both `29 800` and `31 100` are `2` standard deviations away from the mean.
To find the standard deviation, simply subtract the mean `(30 000)` from `31 100`, then divide it by `2` (because of the `2` SDs).`SD=(31 100-30 000)/2``SD=(1 100)/2``SD=550``SD = 550` -
Question 2 of 5
2. Question
`20 000` light bulbs are tested to see how long they last. The collected data is normally distributed. Within how many hours did `68%` of the light bulbs last?
- (29450) and (30550) hours
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Given Values
Mean`= 30 000`Since the given curve is labelled, we can easily see which numbers lie `1` standard deviation above and below the mean.
`68%` of the light bulbs lasted within `29 450` and `30 550` hours.`29 450` and `30 550` hours -
Question 3 of 5
3. Question
`20 000` light bulbs are tested to see how long they last. The collected data is normally distributed. How many light bulbs last less than `29 450` hours?
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Given Values
Mean`= 30 000`We are asked about how many light bulbs last less than `29 450` hours.
First, we find the percentage of these light bulbs.All the data below the mean `(30 000)` is `50%`.
Knowing that `68%` of the data lies `1` SD below and above the mean, we can say that the data between `29 450` and `30 000` is `34%` because it is `1` SD just below the mean.
Compute for the shaded region.`50%-34%``=16%`The percentage of the data below `29 450` is `16%`.Simply get `16%` of all the light bulbs tested.`16/100=0.16``20 000xx0.16=3200`Hence, `3200` of the light bulbs lasted less than `29 450` hours.`3200` -
Question 4 of 5
4. Question
Skull widths of a group of adults are normally distributed with a standard deviation of `52` mm. What is the mean width if `2.5%` of the skull widths are less than `1276` mm?Hint
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Given Values
Standard Deviation`= 52`First, visualize the bell curve for this problem.We know that `50%` of the data lies on the first half of the curve.
Also, remember that `95%` of the data lies `2` standard deviations above and below the mean. This leaves us with `5%` for the rest of the curve, specifically, `2.5%` for each tail outside the `95%`.
Subtract `2.5%` from `50%` to find the percentage of the data between `1276` and the mean.`50%-2.5%=47.5%``47.5%` is actually half of `95%`. This means that there are `2` standard deviations between `1276` and the mean.
Given that the standard deviation is `52` mm, we can easily compute for the mean width.`1276+``52``+` `52``=1380`The mean width is `1380` mm.`1380` -
Question 5 of 5
5. Question
Skull widths of a group of adults are normally distributed with a standard deviation of `52` mm. If there are `10 000` adult skulls, how many skulls have widths between `1224` and `1484` mm?
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Standard Deviation`= 52`Mean `barX= 1380`First, complete the labels of the bell curve by using the mean and standard deviation.For example, start with the mean, `1380` mm. Then add and subtract `52` mm to get the values `1` standard deviation above and below the mean.
Keep adding and subtracting the standard deviation until the labels are completed.Also, shade the region between the values in question, `1224` and `1484`.
The region from the mean to the left has `3` SDs.While the region from the mean to the right has `2` SDs.
Remember that `3` SDs from above and below the mean covers `99.7%` of the data. While `2` SDs cover `95%`.Knowing this, compute for the percentage of the shaded region.`(99.7%)/2+(95%)/2``=49.85%+47.5%``=97.35%`The percentage of the shaded region is `97.35%`Simply get `97.35%` of all the adult skulls.`97.35/100=0.9735``10 000xx0.9735=9735`Hence, `9735` adult skulls have width between `1224` and `1484`.`9735`