Coaching Centre
Applications and quality control
Solving practical problems with normal distributions: percentages and numbers of items, setting cut-offs, and using the mean and standard deviation for quality control in manufacturing.
A bottling plant in Western Sydney fills thousands of $600$ mL bottles every hour. No machine is perfect: some bottles get $603$ mL and some get $597$ mL. Too little and the company breaks consumer law. Too much and it gives away drink for free. The normal distribution tells them exactly how many bottles will be out, and when the machine needs fixing.
A method for any problem
- Write down the mean and standard deviation.
- Find the z-score of each value in the question (or count standard deviations from the mean).
- Sketch a bell curve and shade the region you want.
- Add the pieces: $34$, $13.5$, $2.35$, $0.15$, or $50$ for a whole half.
- If asked "how many", multiply the percentage by the total.
The bottles have volumes normally distributed with mean $600$ mL and $s = 4$ mL. Bottles under $592$ mL are rejected. In a run of $12{,}000$ bottles, how many are rejected?
Working backwards to a cut-off
Sometimes you know the percentage and need the value. Turn the percentage into a z-score using the rule, then use $x = \bar{x} + zs$.
A gym's fitness test scores are normal with mean $50$ and $s = 6$. The top $16\%$ get a free month. What score do you need?
Quality control
Factories don't weigh every product. Instead they take samples and check them against the expected mean and standard deviation. The usual rules are based on the 68–95–99.7 rule:
- A product within $2$ standard deviations of the target is normal. About $95\%$ will be there.
- A product beyond $2$ standard deviations happens only about $5\%$ of the time. Several in a row is a warning sign.
- A product beyond $3$ standard deviations happens only $0.3\%$ of the time. One of these usually means stop the machine and check it.
Set the machine's target and its standard deviation, then press Run 100 bottles. The histogram and the reject count build up. Bottles below the reject line turn red.
Press Machine fault: the mean starts to drift. How many bottles does it take before you'd notice? What standard deviation do you need so that fewer than $1$ bottle in $40$ is rejected?
Your turn
Bags of rice are labelled $1$ kg. The masses are normal with mean $1{,}010$ g and $s = 5$ g. (a) What percentage of bags are under $1{,}000$ g? (b) A bag of $1{,}027$ g is found. Should the machine be checked?
Show the answer
(a) $z = \dfrac{1000 - 1010}{5} = -2$. Below $z = -2$ is $2.5\%$.
(b) $z = \dfrac{1027 - 1010}{5} = \dfrac{17}{5} = 3.4$. That's more than $3$ standard deviations above the mean, which should happen less than $0.15\%$ of the time. The machine should be checked.
Cans of soup have mass normally distributed with mean $410$ g and $s = 3$ g. What percentage of cans weigh more than $416$ g?
$z = \dfrac{416 - 410}{3} = 2$. Above $z = 2$ is $5 \div 2 = 2.5\%$.
In a run of $20{,}000$ cans from the same machine, how many weigh more than $416$ g?
$2.5\%$ of $20{,}000$: $0.025 \times 20{,}000 = 500$ cans.
Light globes last a mean of $1{,}200$ hours with $s = 100$ hours, normally distributed. What percentage last between $1{,}100$ and $1{,}300$ hours?
$1{,}100$ and $1{,}300$ are $z = -1$ and $z = 1$. Within $1$ standard deviation is $68\%$.
Using the same globes, what percentage last more than $1{,}000$ hours?
$1{,}000$ is $z = \dfrac{1000 - 1200}{100} = -2$. Above $z = -2$ is $47.5 + 50 = 97.5\%$.
Scores are normal with mean $40$ and $s = 9$. The top $2.5\%$ of scores are above what value?
The top $2.5\%$ lies above $z = 2$. $x = 40 + 2 \times 9 = 58$.
A machine fills bags with mean $250$ g and $s = 2$ g. Which bag should make an inspector check the machine?
z-scores: $249 \to -0.5$, $253 \to 1.5$, $257 \to 3.5$, $246.5 \to -1.75$. Only $257$ g is beyond $3$ standard deviations, which should almost never happen.
Over a week, $12\%$ of bottles from a machine are more than $2$ standard deviations from the target. What does this suggest?
The 68–95–99.7 rule predicts about $5\%$ beyond $\pm 2s$. $12\%$ is more than double that, so the mean or the spread has probably changed.
Bottles have volume mean $600$ mL and $s = 4$ mL. How many of $12{,}000$ bottles contain between $596$ mL and $608$ mL?
$z = -1$ to $z = 2$: $34 + 34 + 13.5 = 81.5\%$. Then $0.815 \times 12{,}000 = 9{,}780$ bottles.
Exam style: The time taken for a Sydney Trains service from Central to Hornsby is normal with mean $40$ minutes. $84\%$ of trips take less than $44$ minutes. Find the median trip time.
For normal data the median equals the mean, so the median is $40$ minutes. (The $84\%$ tells you $44 = \bar{x} + s$, so $s = 4$, but that isn't needed for the median.)
Exam style: A machine fills boxes of cereal with mean $500$ g. Inspectors find that $2.5\%$ of boxes weigh less than $488$ g. Assuming the masses are normally distributed, find the standard deviation.
The bottom $2.5\%$ lies below $z = -2$. So $488 = 500 - 2s$, giving $2s = 12$ and $s = 6$ g.