The Normal Distribution (HSC SSCE Mathematics Standard): Revision Notes

Worked Example: Heights of Children

Question: The heights of children are normally distributed with a mean of $150$ cm and a standard deviation of $5$ cm. What percentage of children have heights between $145$ cm and $155$ cm?

Solution:

We need to find the z-scores for both boundary values.

For the lower boundary ($145$ cm):

$z = \frac{x - \bar{x}}{s}$

$z = \frac{145 - 150}{5}$

$z = \frac{-5}{5}$

$z = -1$

For the upper boundary ($155$ cm):

$z = \frac{x - \bar{x}}{s}$

$z = \frac{155 - 150}{5}$

$z = \frac{5}{5}$

$z = 1$

The heights between $145$ cm and $155$ cm correspond to z-scores between $-1$ and $1$. According to the 68-95-99.7 rule, 68% of values fall within this range.

Answer: 68% of children have heights between 145 cm and 155 cm.

Worked Example: Mass of Jam Jars

Question: The mass of $700$ jars of jam is normally distributed with a mean of $420$ grams and a standard deviation of $3$ grams.

a) What percentage of the jars have a mass:

• i) more than $426$ grams?
• ii) less than $423$ grams?

b) What percentage of jars are predicted to have a mass between $414$ and $429$ grams?

Solution:

Part a(i): Finding the percentage more than 426 grams

First, calculate the z-score for $426$ grams:

$z = \frac{x - \bar{x}}{s}$

$z = \frac{426 - 420}{3}$

$z = \frac{6}{3}$

$z = 2$

The z-score of $2$ tells us that $426$ grams is two standard deviations above the mean.

According to the 68-95-99.7 rule, 95% of values fall between z-scores of $-2$ and $2$. This means:

Percentage outside this range $= 100\% - 95\% = 5\%$

Due to symmetry, half of this 5% is above a z-score of $2$, and half is below $-2$.

$5\% \div 2 = 2.5\%$

Answer: 2.5% of jars have a mass more than 426 grams.

Part a(ii): Finding the percentage less than 423 grams

Calculate the z-score for $423$ grams:

$z = \frac{x - \bar{x}}{s}$

$z = \frac{423 - 420}{3}$

$z = \frac{3}{3}$

$z = 1$

The z-score of $1$ means $423$ grams is one standard deviation above the mean.

From the 68-95-99.7 rule, 68% of values fall between z-scores of $-1$ and $1$. Due to symmetry, half of this percentage falls between the mean (z-score of $0$) and a z-score of $1$:

$68\% \div 2 = 34\%$

This 34% represents values between the mean and $423$ grams.

Since 50% of values fall below the mean, we add:

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

Answer: 84% of jars have a mass less than 423 grams.

Part b: Finding the percentage between 414 and 429 grams

Calculate the z-score for $414$ grams:

$z = \frac{x - \bar{x}}{s}$

$z = \frac{414 - 420}{3}$

$z = \frac{-6}{3}$

$z = -2$

Calculate the z-score for $429$ grams:

$z = \frac{x - \bar{x}}{s}$

$z = \frac{429 - 420}{3}$

$z = \frac{9}{3}$

$z = 3$

We need the percentage of values between z-scores of $-2$ and $3$.

From the mean to a z-score of $-2$ (using symmetry):

$95\% \div 2 = 47.5\%$

From the mean to a z-score of $3$ (using symmetry):

$99.7\% \div 2 = 49.85\%$

Total percentage between $-2$ and $3$:

$47.5\% + 49.85\% = 97.35\%$

Answer: 97.35% of jars are predicted to have a mass between 414 and 429 grams.