The 68–95–99.7 Rule 1

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Year 10

Normal Distribution

The 68–95–99.7 Rule

The 68–95–99.7 Rule 1

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Question 1 of 4

1. Question
Decking screws are produced by a machine and the lengths are normally distributed with a mean of $6$ $6$ cm and a standard deviation of $0.02$ $0.02$ cm. What interval does the following percentages cover?

$(a) 68 %$ $\text((a)) 68%$

$(b) 95 %$ $\text((b)) 95%$

$(c) 99.7 %$ $\text((c)) 99.7%$

$(d)$ $\text((d))$ What is the percentage of the lengths that are more than $3$ $3$ standard deviations away from the mean?
- $(a)$ $\text((a))$ (5.98) to (6.02) cm
  $(b)$ $\text((b))$ (5.96) to (6.04) cm
  $(c)$ $\text((c))$ (5.94) to (6.06) cm
  $(d)$ $\text((d))$ (0.15) $%$ $%$
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Mean $\bar{X} = 6$ $barX= 6$ cm

Standard Deviation $= 0.02$ $=0.02$ cm

For example, start with the mean, $6$ $6$ cm. Then add and subtract $0.02$ $0.02$ cm 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.

Remember that $68 %$ $68%$ of the data lies $1$ $1$ SD below and above the mean.

Hence, $68 %$ $68%$ of the decking screws are $5.98$ $5.98$ to $6.02$ $6.02$ cm long.

Remember that $95 %$ $95%$ of the data lies $2$ $2$ SDs below and above the mean.

Hence, $95 %$ $95%$ of the decking screws are $5.96$ $5.96$ to $6.04$ $6.04$ cm long.

Remember that $99.7 %$ $99.7%$ of the data lies $3$ $3$ SDs below and above the mean.

Hence, $99.7 %$ $99.7%$ of the decking screws are $5.94$ $5.94$ to $6.06$ $6.06$ cm long.

Here, we are asked to find the percentage of the lengths that are more than $3$ $3$ SDs from the mean.

Subtract $99.7 %$ $99.7%$ from $100 %$ $100%$ , then divide it by $2$ $2$ to get the percentage.

$\frac{100 % - 99.7 %}{2} = \frac{0.3 %}{2} = 0.15 %$ $(100%-99.7%)/2=(0.3%)/2=0.15%$

Hence, $0.15 %$ $0.15%$ of the lengths are more than $3$ $3$ standard deviations from the mean.

$(a) 5.98 to 6.02$ $\text((a)) 5.98 \text(to) 6.02$

$(b) 5.96 to 6.04$ $\text((b)) 5.96 \text(to) 6.04$

$(c) 5.94 to 6.06$ $\text((c)) 5.94 \text(to) 6.06$

$(d) 0.15 %$ $\text((d)) 0.15%$
Question 2 of 4

2. Question
A series of science test scores from students have a mean of $52.8$ $52.8$ and a standard deviation of $14.4$ $14.4$ . How many scores are more than $1$ $1$ standard deviation away from the mean?

$62, 53, 47, 29, 55, 72, 71, 52, 38, 49$ $62, 53, 47, 29, 55, 72, 71, 52, 38, 49$
- 1. $4$ $4$
- 2. $6$ $6$
- 3. $3$ $3$
- 4. $7$ $7$
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Mean $\bar{X} = 52.8$ $barX= 52.8$ kg

Standard Deviation $= 14.4$ $=14.4$ kg

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

For example, start with the mean, $52.8$ $52.8$ . Then add and subtract $14.4$ $14.4$ 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 how many scores are more than $1$ $1$ SD away from the mean.

This means we are looking at both sides of 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.

$4$ $4$
Question 3 of 4

3. Question
A factory makes alloy wheels with a mean diameter of $406$ $406$ mm and a standard deviation of $2$ $2$ mm. The diameters are normally distributed. Diameters more than $3$ $3$ standard deviations away from the mean get rejected. Which alloy wheels will be rejected if their diameters are:

$398$ $398$ , $403$ $403$ , $407$ $407$ , $411$ $411$ , $412.5$ $412.5$
- (398) mm and (412.5) mm
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Given Values

Mean $\bar{X} = 406$ $barX= 406$ mm

Standard Deviation $= 2$ $=2$ mm

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

For example, start with the mean, $406$ $406$ mm. Then add and subtract $2$ $2$ 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.

We are asked about how many wheels measure more than $3$ $3$ SDs away from the mean.

This means we are looking at both sides of the mean.

Now, from the diameters given, list down those that are less than $400$ $400$ or greater than $412$ $412$ .

Hence, the allow wheels with diameters $398$ $398$ mm and $412.5$ $412.5$ mm get rejected.

$398$ $398$ mm and $412.5$ $412.5$ mm
Question 4 of 4

4. Question
Brand $X$ $X$ breakfast cereal is branded with a weight of $700$ $700$ g. The actual weights were found to be normally distributed with a mean of $690$ $690$ g and a standard deviation of $10$ $10$ g. What percent of packets would you expect to be below the labeled weight?
- (84) $%$ $%$
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Mean $\bar{X} = 690$ $barX= 690$ g

Standard Deviation $= 10$ $=10$ g

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

For example, start with the mean, $690$ $690$ g. Then add and subtract $10$ $10$ 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 packets that weigh less than the labeled weight, which is $700$ $700$ g.

In the curve, this means we are looking for the percentage from $700$ $700$ to the left.

All the data below the mean $(690)$ $(690)$ 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 $690$ $690$ and $700$ $700$ is $34 %$ $34%$ because it is $1$ $1$ SD just above the mean.

Compute for the final percentage.

$50 % + 34 %$ $50%+34%$ $= 84 %$ $=84%$

The percentage of the packets that weigh less than the labeled weight is $84 %$ $84%$ .

$84 %$ $84%$

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1. Question

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4. Question

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