The 68–95–99.7 Rule 3

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Year 10

Normal Distribution

The 68–95–99.7 Rule

The 68–95–99.7 Rule 3

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Question 1 of 5

1. Question
$20000$ $20 000$ light bulbs are tested to see how long they last. The collected data is normally distributed. If $95 %$ $95%$ of the light bulbs lasted within $29800$ $29 800$ and $31100$ $31 100$ hours, what is the standard deviation? The mean is $30, 000$ $30,000$ .
- 1. $500$ $500$
- 2. $610$ $610$
- 3. $480$ $480$
- 4. $550$ $550$
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Mean $= 30000$ $= 30 000$

This means that both $29800$ $29 800$ and $31100$ $31 100$ are $2$ $2$ standard deviations away from the mean.

To find the standard deviation, simply subtract the mean $(30000)$ $(30 000)$ from $31100$ $31 100$ , then divide it by $2$ $2$ (because of the $2$ $2$ SDs).

$S D = \frac{31100 - 30000}{2}$ $SD=(31 100-30 000)/2$

$S D = \frac{1100}{2}$ $SD=(1 100)/2$

$S D = 550$ $SD=550$

$S D = 550$ $SD = 550$
Question 2 of 5

2. Question
$20000$ $20 000$ light bulbs are tested to see how long they last. The collected data is normally distributed. Within how many hours did $68 %$ $68%$ of the light bulbs last?
- (29450) and (30550) hours
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Given Values

Mean $= 30000$ $= 30 000$

Since the given curve is labelled, we can easily see which numbers lie $1$ $1$ standard deviation above and below the mean.

$68 %$ $68%$ of the light bulbs lasted within $29450$ $29 450$ and $30550$ $30 550$ hours.

$29450$ $29 450$ and $30550$ $30 550$ hours
Question 3 of 5

3. Question
$20000$ $20 000$ light bulbs are tested to see how long they last. The collected data is normally distributed. How many light bulbs last less than $29450$ $29 450$ hours?
- 1. $4000$ $4000$
- 2. $6400$ $6400$
- 3. $3200$ $3200$
- 4. $6800$ $6800$
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Given Values

Mean $= 30000$ $= 30 000$

We are asked about how many light bulbs last less than $29450$ $29 450$ hours.

First, we find the percentage of these light bulbs.

All the data below the mean $(30000)$ $(30 000)$ is $50 %$ $50%$ .

Knowing that $68 %$ $68%$ of the data lies $1$ $1$ SD below and above the mean, we can say that the data between $29450$ $29 450$ and $30000$ $30 000$ is $34 %$ $34%$ because it is $1$ $1$ SD just below the mean.

Compute for the shaded region.

$50 % - 34 %$ $50%-34%$ $= 16 %$ $=16%$

The percentage of the data below $29450$ $29 450$ is $16 %$ $16%$ .

Simply get $16 %$ $16%$ of all the light bulbs tested.

$\frac{16}{100} = 0.16$ $16/100=0.16$

$20000 \times 0.16 = 3200$ $20 000xx0.16=3200$

Hence, $3200$ $3200$ of the light bulbs lasted less than $29450$ $29 450$ hours.

$3200$ $3200$
Question 4 of 5

4. Question
Skull widths of a group of adults are normally distributed with a standard deviation of $52$ $52$ mm. What is the mean width if $2.5 %$ $2.5%$ of the skull widths are less than $1276$ $1276$ mm?
- 1. $1432$ $1432$
- 2. $1224$ $1224$
- 3. $1328$ $1328$
- 4. $1380$ $1380$
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Given Values

Standard Deviation $= 52$ $= 52$

First, visualize the bell curve for this problem.

We know that $50 %$ $50%$ of the data lies on the first half of the curve.

Also, remember that $95 %$ $95%$ of the data lies $2$ $2$ standard deviations above and below the mean. This leaves us with $5 %$ $5%$ for the rest of the curve, specifically, $2.5 %$ $2.5%$ for each tail outside the $95 %$ $95%$ .

Subtract $2.5 %$ $2.5%$ from $50 %$ $50%$ to find the percentage of the data between $1276$ $1276$ and the mean.

$50 % - 2.5 % = 47.5 %$ $50%-2.5%=47.5%$

$47.5 %$ $47.5%$ is actually half of $95 %$ $95%$ . This means that there are $2$ $2$ standard deviations between $1276$ $1276$ and the mean.

Given that the standard deviation is $52$ $52$ mm, we can easily compute for the mean width.

$1276 +$ $1276+$ $52$ $52$ $+$ $+$ $52$ $52$ $= 1380$ $=1380$

The mean width is $1380$ $1380$ mm.

$1380$ $1380$
Question 5 of 5

5. Question
Skull widths of a group of adults are normally distributed with a standard deviation of $52$ $52$ mm. If there are $10000$ $10 000$ adult skulls, how many skulls have widths between $1224$ $1224$ and $1484$ $1484$ mm?
- 1. $9735$ $9735$
- 2. $8545$ $8545$
- 3. $9950$ $9950$
- 4. $7400$ $7400$
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Given Values

Standard Deviation $= 52$ $= 52$

Mean $¯ ¯ ¯ X = 1380$ $barX= 1380$

First, complete the labels of the bell curve by using the mean and standard deviation.

For example, start with the mean, $1380$ $1380$ mm. Then add and subtract $52$ $52$ mm to get the values $1$ $1$ 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, $1224$ $1224$ and $1484$ $1484$ .

The region from the mean to the left has $3$ $3$ SDs.

While the region from the mean to the right has $2$ $2$ SDs.

Remember that $3$ $3$ SDs from above and below the mean covers $99.7 %$ $99.7%$ of the data. While $2$ $2$ SDs cover $95 %$ $95%$ .

Knowing this, compute for the percentage of the shaded region.

$\frac{99.7 %}{2} + \frac{95 %}{2}$ $(99.7%)/2+(95%)/2$

$= 49.85 % + 47.5 %$ $=49.85%+47.5%$

$= 97.35 %$ $=97.35%$

The percentage of the shaded region is $97.35 %$ $97.35%$

Simply get $97.35 %$ $97.35%$ of all the adult skulls.

$\frac{97.35}{100} = 0.9735$ $97.35/100=0.9735$

$10000 \times 0.9735 = 9735$ $10 000xx0.9735=9735$

Hence, $9735$ $9735$ adult skulls have width between $1224$ $1224$ and $1484$ $1484$ .

$9735$ $9735$

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