Probabilities for the Sample Mean
Four lessons of build-up, for this. You know the sampling distribution is centred on $\mu$ , that its standard deviation is $\sigma/\sqrt{n}$ , and that for $n \geq 30$ it is approximately normal. Put those together and you can finally answer the question the whole topic exists for: how likely is my sample mean to be close to the truth?
Two numbers, before you read on. Most people put them much closer together than they are.
A machine fills bags with mean $500$ g and standard deviation $80$ g. (a) How likely is the mean of $64$ bags to land between $490$ g and $510$ g? (b) How likely is a single bag to land between $490$ g and $510$ g?
- Find $\text{sd}(\bar{X}) = \dfrac{\sigma}{\sqrt{n}}$ , and check $n \geq 30$ . Without that check you have no right to the normal curve at all.
- Standardise each bound and read the area. Convert the given values of $\bar{x}$ into $z$ , then use the empirical rule to turn the $z$ range into a probability.
- Standardise a sample mean with $z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ .
- Use the empirical rule to estimate the probability that $\bar{X}$ lies within given bounds .
- Handle the three shapes: between two bounds, above a bound, below a bound.
- Explain why the same bounds are far more likely for a mean than for one observation.
By the central limit theorem, for $n \geq 30$ the sample mean is approximately $N\!\left(\mu, \dfrac{\sigma^2}{n}\right)$ . To find a probability you convert $\bar{x}$ to a $z$ value in the usual way, using that distribution's standard deviation:
$$z = \frac{\bar{x} - \mu}{\dfrac{\sigma}{\sqrt{n}}}$$Then read the area from the empirical rule:
| $z$ range | area inside | area outside | each tail |
|---|---|---|---|
| $-1$ to $1$ | $68\%$ | $32\%$ | $16\%$ |
| $-2$ to $2$ | $95\%$ | $5\%$ | $2.5\%$ |
| $-3$ to $3$ | $99.7\%$ | $0.3\%$ | $0.15\%$ |
The hook, worked. Bags with $\mu = 500$ , $\sigma = 80$ , and $n = 64$ :
- $\text{sd}(\bar{X}) = \dfrac{80}{\sqrt{64}} = \dfrac{80}{8} = 10$ g, and $n = 64 \geq 30$ , so the theorem applies.
- $490$ standardises to $z = \dfrac{490 - 500}{10} = -1$ , and $510$ to $z = +1$ .
- So $P(490 < \bar{X} < 510) = P(-1 < z < 1) = \mathbf{68\%}$ .
To find a probability for a sample mean, first check n ≥ 30 and compute sd(X̄) = σ/√n, then standardise each bound with z = (x̄ − μ) ÷ (σ/√n) and read the area from the empirical rule: 68% within ±1, 95% within ±2, 99.7% within ±3.
Pause, copy the $z$ formula for a sample mean with the $\sigma/\sqrt{n}$ clearly in the denominator, and copy the empirical-rule table including the tail column.
Quick check: With $\mu = 500$ , $\sigma = 80$ and $n = 64$ , what is $z$ for $\bar{x} = 520$ ?
Every question of this kind is one of three shapes, and each is the same two moves followed by a different reading. Staying with the bags, where $\text{sd}(\bar{X}) = 10$ :
| Asked | Standardised | Read as | Answer |
|---|---|---|---|
| $P(480 < \bar{X} < 520)$ | $P(-2 < z < 2)$ | the area inside $\pm 2$ | $95\%$ |
| $P(\bar{X} > 520)$ | $P(z > 2)$ | half of the $5\%$ outside | $2.5\%$ |
| $P(\bar{X} < 480)$ | $P(z < -2)$ | the other half, by symmetry | $2.5\%$ |
Two facts do all the work in the second and third rows: the curve is symmetric , and the total area is $1$ . A one-sided question is always the inside area subtracted from $1$ , then halved if symmetry applies.
Now the comparison the hook was really about. Ask the same question about a single bag rather than the mean of $64$ :
| Between $490$ g and $510$ g | divide by | $z$ range | probability |
|---|---|---|---|
| the mean of $64$ bags | $\sigma/\sqrt{n} = 10$ | $-1$ to $1$ | about $68\%$ |
| one single bag | $\sigma = 80$ | $-0.125$ to $0.125$ | about $10\%$ |
Between two bounds, read the inside area. Above or below a bound, subtract the inside area from 1 and halve it if the curve's symmetry applies. The same bounds are far more likely for a sample mean than for one observation, because the mean's spread is σ/√n rather than σ.
Pause, copy the three shapes with their readings, and copy the two-row comparison showing $68\%$ against $10\%$ for the same interval.
True or false: Because the population has $\sigma = 80$ , the probability that the mean of $64$ bags lies within $10$ g of $500$ g is the same as the probability that one bag does.
Worked examples · 3 in a row, reveal as you go
Test scores have $\mu = 72$ and $\sigma = 15$ . A random sample of $25$ scores is taken. Estimate $P(66 < \bar{X} < 78)$ . (3 marks)
Bags have $\mu = 500$ g and $\sigma = 80$ g. For a sample of $64$ bags, estimate the probability that the sample mean exceeds $520$ g. (2 marks)
For the same bags, $200$ separate samples of $64$ are taken and each sample mean recorded. (a) About how many of those $200$ means would you expect to fall between $470$ g and $530$ g? (b) About how many below $480$ g? (3 marks)
Complete: To standardise a sample mean, divide by , and the area within $\pm 2$ of the mean is %.
Misconceptions to fix · the 3 traps that cost marks
True or false: If $5\%$ of the area lies outside $z = \pm 2$ , then $P(\bar{X} > \mu + 2\,\text{sd}) = 5\%$ .
Activities · practice with the ideas
A population has $\mu = 200$ and $\sigma = 24$ . For $n = 36$ , find $\text{sd}(\bar{X})$ , then estimate $P(196 < \bar{X} < 204)$ .
For the same population and sample size, estimate $P(\bar{X} > 208)$ .
Explain in one sentence why $P(196 < \bar{X} < 204)$ is much larger than the probability that a single observation falls between $196$ and $204$ .
$400$ samples of size $36$ are taken from that population. About how many sample means would you expect below $192$ ?
A student standardises a sample mean by dividing by $\sigma$ . Describe the effect on their answer, and how you would spot it in their working.
Which does NOT belong? Steps in estimating $P(a < \bar{X} < b)$ :
Earlier you estimated two probabilities for the same interval, $490$ g to $510$ g.
For the mean of $64$ bags it is about $68\%$ , because $\text{sd}(\bar{X}) = 80/\sqrt{64} = 10$ and the interval is exactly one standard deviation either side. For a single bag it is about $10\%$ , because that same interval is only $\pm 0.125$ of the population's own standard deviation of $80$ . Nearly seven times the probability, from the identical population and the identical bounds.
That gap is the whole of Module 10's second half in one number. Lesson 21 said a sample can stand in for a population it cannot measure. Lesson 22 showed sample means vary. Lesson 23 gave that variation a distribution, Lesson 24 gave it a shape and a width, and this lesson turned it into a probability you can act on. ME1-12-06 is complete.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A population has $\mu = 150$ and $\sigma = 20$ . A random sample of $100$ is taken. (a) Find $\text{sd}(\bar{X})$ . (b) Estimate $P(146 < \bar{X} < 154)$ , justifying your use of the normal approximation. (3 marks)
Q2. For the same population and sample size, estimate $P(\bar{X} > 156)$ . Show how you obtain a one-tailed probability from the empirical rule. (3 marks)
Q3. Still with $\mu = 150$ , $\sigma = 20$ , $n = 100$ . (a) Five hundred separate samples are taken and each sample mean recorded. About how many would you expect below $144$ ? (b) A student standardises using $\sigma = 20$ instead of $\text{sd}(\bar{X})$ . State the factor by which their $z$ values are wrong, and say whether their probabilities come out too large or too small. (c) Explain what would change in your answer to (a) if the sample size were $9$ rather than $100$ . (4 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\text{sd}(\bar{X}) = \dfrac{24}{\sqrt{36}} = 4$ . Then $196$ and $204$ give $z = -1$ and $z = 1$ , so $P \approx 68\%$ .
2. $z = \dfrac{208 - 200}{4} = 2$ . The area inside $\pm 2$ is $95\%$ , so $5\%$ lies outside, and by symmetry one tail is $\mathbf{2.5\%}$ .
3. Because the sample mean varies far less than a single observation: its standard deviation is $\sigma/\sqrt{n} = 4$ rather than $\sigma = 24$ . The same interval is therefore $\pm 1$ standard deviation for the mean but only about $\pm 0.17$ for one observation.
4. $z = \dfrac{192 - 200}{4} = -2$ , so $P(\bar{X} < 192) = 2.5\%$ , and $0.025 \times 400 = \mathbf{10}$ sample means.
5. Dividing by $\sigma$ instead of $\sigma/\sqrt{n}$ makes every $z$ value $\sqrt{n} = 6$ times too small, so the probability of being close to $\mu$ comes out far too low. You would spot it in the working as a denominator with no $\sqrt{n}$ in it, and in the answer as an implausibly small probability for a large sample.
Q1 (3 marks): (a) $\text{sd}(\bar{X}) = \dfrac{20}{\sqrt{100}} = \dfrac{20}{10} = 2$ [1]. (b) $146$ gives $z = \dfrac{146 - 150}{2} = -2$ and $154$ gives $z = 2$ [1], so $P(-2 < z < 2) \approx 95\%$ . The normal approximation is justified by the central limit theorem because $n = 100 \geq 30$ , and no assumption about the population's shape is needed [1].
Q2 (3 marks): $z = \dfrac{156 - 150}{2} = 3$ [1]. The empirical rule gives $99.7\%$ inside $\pm 3$ , so $0.3\%$ lies outside [1]. The curve is symmetric, so one tail is half of that: $P(\bar{X} > 156) \approx \mathbf{0.15\%}$ [1].
Q3 (4 marks): (a) $144$ gives $z = -3$ , so $P(\bar{X} < 144) \approx 0.15\%$ , and $0.0015 \times 500 = 0.75$ , so about $1$ of the $500$ sample means [1]. (b) Dividing by $20$ instead of $2$ makes every $z$ value $10$ times too small, that is $\sqrt{n}$ times [1]. A smaller $z$ puts the bound nearer the centre, so the tail probabilities come out far too large and the "within bounds" probabilities far too small [1]. (c) With $n = 9$ , $\text{sd}(\bar{X})$ would be $\dfrac{20}{3} \approx 6.67$ rather than $2$ , so $144$ would be less than one standard deviation below the mean and the probability would be much larger. More importantly $n = 9 < 30$ , so the central limit theorem no longer licenses the normal approximation, and the estimate would only be defensible if the population were already known to be roughly normal [1].
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