The 68–95–99.7 Rule 2

Years

Year 10

Normal Distribution

The 68–95–99.7 Rule

The 68–95–99.7 Rule 2

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Question 1 of 4

1. Question
Tubs of natural yogurt are labeled as weighing $1000$ $1000$ g. Surveys found the weights to be normally distributed with a mean of $1022$ $1022$ g and a standard deviation of $11$ $11$ g. What percentage of the tubs contain more than the labeled weight?
- (97.5, 97.50) $%$ $%$
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Mean $¯ ¯ ¯ X = 1022$ $barX= 1022$ g

Standard Deviation $= 11$ $= 11$ g

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

For example, start with the mean, $1022$ $1022$ g. Then add and subtract $11$ $11$ g 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.

We are asked about the percentage of tubs that weigh more than the labeled weight, which is $1000$ $1000$ g.

In the curve, this means we are looking for the percentage from $1000$ $1000$ onwards.

All the data above the mean $(1022)$ $(1022)$ is $50 %$ $50%$

Knowing that $95 %$ $95%$ of the data lies $2$ $2$ SDs below and above the mean, we can say that the data between $1000$ $1000$ and $1022$ $1022$ is $2$ $2$ SDs just below the mean.

Compute for the final percentage.

$50 % + \frac{95 %}{2}$ $50%+(95%)/2$

$= 50 % + 47.5 %$ $=50%+47.5%$

$= 97.5 %$ $=97.5%$

The percentage of the tubs that weigh more than the labeled weight is $97.5 %$ $97.5%$ .

$97.5 %$ $97.5%$
Question 2 of 4

2. Question
Tubs of natural yogurt are labeled as weighing $1000$ $1000$ g. Surveys found the weights to be normally distributed with a mean of $1022$ $1022$ g and a standard deviation of $11$ $11$ g. If there are $4000$ $4000$ tubs included in the survey, how many contain less than the labelled weight?
- 1. $100$ $100$
- 2. $75$ $75$
- 3. $200$ $200$
- 4. $250$ $250$
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Given Values

Mean $¯ ¯ ¯ X = 1022$ $barX= 1022$ g

Standard Deviation $= 11$ $= 11$ g

We are asked about how many tubs weighed less than $1000$ $1000$ g.

First, we find the percentage of these tubs.

All the data below the mean $(1022)$ $(1022)$ is $50 %$ $50%$ .

Knowing that $95 %$ $95%$ of the data lies $2$ $2$ SDs below and above the mean, we can say that the data between $1000$ $1000$ and $1022$ $1022$ is $2$ $2$ SDs just below the mean.

Now, from the scores given, list down those that are less than $39.4$ $39.4$ or greater than $66.2$ $66.2$ .

$29, 72, 71, 38$ $29, 72, 71, 38$

Hence, there are $4$ $4$ scores that are more than $1$ $1$ standard deviation away from the mean.

Compute for the final percentage.

$50 % - \frac{95 %}{2}$ $50%-(95%)/2$

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

$= 2.5 %$ $=2.5%$

The percentage of the tubs that weigh less than the labeled weight is $2.5 %$ $2.5%$ .

Simply get $2.5 %$ $2.5%$ of all the tubs surveyed.

$\frac{2.5}{100} = 0.025$ $2.5/100=0.025$

$4000 \times 0.025 = 100$ $4000xx0.025=100$

Hence, $100$ $100$ of the tubs weighed less than $1000$ $1000$ g.

$100$ $100$
Question 3 of 4

3. Question
There are packets of sugar each labelled as $2$ $2$ kg. When the weights of $500$ $500$ of these packets were checked, they were found to be normally distributed with a mean of $2.025$ $2.025$ kg and a standard deviation of $0.025$ $0.025$ kg. What percentage of these packets weigh less than the labelled weight?
- (16) $%$ $%$
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Given Values

Mean $¯ ¯ ¯ X = 2.025$ $barX= 2.025$ kg

Standard Deviation $= 0.025$ $=0.025$ kg

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

For example, start with the mean, $2.025$ $2.025$ kg. Then add and subtract $0.025$ $0.025$ kg 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.

In the curve, this means we are looking for the percentage from $2.000$ $2.000$ to the left.

All the data below the mean $(2.025)$ $(2.025)$ 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 $2.000$ $2.000$ and $2.025$ $2.025$ is $34 %$ $34%$ because it is $1$ $1$ SD just below the mean.

Compute for the final percentage.

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

The percentage of the packets that weigh less than the labelled weight is $16 %$ $16%$ .

$16 %$ $16%$
Question 4 of 4

4. Question
There are packets of sugar each labelled as $2$ $2$ kg. When the weights of $500$ $500$ of these packets were checked, they were found to be normally distributed with a mean of $2.025$ $2.025$ kg and a standard deviation of $0.025$ $0.025$ kg. The packaging machine is reset so that the average weight now is $2.050$ $2.050$ . What percentage is now less than the labelled weight?
- (2.5, 2.50) $%$ $%$
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Given Values

Mean $¯ ¯ ¯ X = 2.050$ $barX= 2.050$ kg

Standard Deviation $= 0.025$ $=0.025$ kg

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

For example, start with the mean, $2.050$ $2.050$ kg. Then add and subtract $0.025$ $0.025$ kg 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.

We are asked about the percentage of packets that weigh less than the labelled weight, which is $2$ $2$ kg.

In the curve, this means we are looking for the percentage from $2.000$ $2.000$ to the left.

All the data below the mean $(2.050)$ $(2.050)$ is $50 %$ $50%$ .

Knowing that $95 %$ $95%$ of the data lies $2$ $2$ SD below and above the mean, we can say that the data between $2.000$ $2.000$ and $2.050$ $2.050$ is $47.5 %$ $47.5%$ because it is $2$ $2$ SD just below the mean.

Compute for the final percentage.

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

The percentage of the packets that weigh less than the labelled weight is $2.5 %$ $2.5%$ , now that the machine is reset.

$2.5 %$ $2.5%$

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