Chapter 13: Review

The chapter problem, solved

The supplier makes two claims on the packet of 20 seeds.

Each seed germinates with probability 0.8, and the seeds germinate independently.
The height of a mature plant is normally distributed with mean 45 cm and standard deviation 6 cm.

The gardener planted all 20 seeds, and 13 of them germinated.

Step 1. Identify the two quantities

The number that germinate is a count. It takes whole-number values from 0 to 20, and the gardener can list every possible value.

The height of a plant is a measurement. It can take any value in a range, and the gardener cannot list every possible height.

The two quantities need different models.

Step 2. The number that germinate

Let $X$ be the number that germinate. There are $n = 20$ independent trials, each seed either germinates or does not, and the probability $p = 0.8$ is the same for every seed. So

X \sim B(20, 0.8).

Find $P(X \le 13)$ using the complement and the upper tail:

P(X \le 13) = 1 - P(X \ge 14) = 1 - \left(P(X = 14) + \dots + P(X = 20)\right).

The sum of the terms from 14 to 20 is $0.9133$ (4 d.p.), so

P(X \le 13) = 1 - 0.9133 = 0.0867 \ (4 \text{ d.p.}).

The expected number is $np = 16$ , so 13 is below the centre. A result of 13 or fewer germinations happens in about 8.7% of packets of this kind. It is unusual, but it is not extremely rare. Deciding whether the result is strong evidence against the claim needs the hypothesis-testing tools of the next chapter.

Step 3. The height of a plant

Let $Y$ be the height of a mature plant. Then $Y \sim N(45, 6^2)$ cm. Find the probability that a plant is taller than 52 cm. Standardise:

z = \frac{52 - 45}{6} = 1.1667 \ (4 \text{ d.p.}).

The table gives $\Phi(1.2) = 0.8849$ and $\Phi(1.1) = 0.8643$ , so $\Phi(1.1667) \approx 0.8783$ . Then

P(Y > 52) = 1 - \Phi(1.1667) \approx 1 - 0.8783 = 0.1217.

About 12.2% of mature plants should be taller than 52 cm.

Step 4. Answer the two questions

About 8.7% of packets of 20 seeds produce 13 or fewer germinations when $p = 0.8$ . The gardener's result is on the low side of the model, but it is not a rare event.
About 12.2% of mature plants should be taller than 52 cm under the height model.

The answers use two different models, and each model rests on assumptions. The germination claim assumes independence. If the seeds share a packet, then a storage problem would affect them together, and the trials would not be independent. The height claim assumes the heights are symmetric about 45 cm with no outliers. The gardener should check both assumptions against the data before trusting the models.

Mixed practice — choose the method first

For each question, say which model or method you will use before you calculate: the binomial formula, an at-most or at-least sum, the standard Normal table, the Normal approximation with the continuity correction, a point of inflection, a binomial coefficient, or the correlation and regression from Chapter 11.

Each of 20 seeds germinates with probability 0.8, independently. The random variable $X$ is the number that germinate. Find $P(X \le 13)$ .
The height of a mature sunflower is modelled as $X \sim N(45, 6^2)$ cm. Find $P(X > 52)$ .
The random variable $X \sim B\left(10, \tfrac13\right)$ . Find $P(X \le 2)$ .
A measurement is modelled as $X \sim N(70, 8^2)$ . Find $P(62 < X < 78)$ .
The random variable $X \sim N(100, 15^2)$ . Find the value $a$ such that $P(X < a) = 0.90$ .
The pairs of values $(1, 2)$ , $(2, 3)$ , $(3, 5)$ , $(4, 4)$ and $(5, 6)$ are plotted. Work out Pearson product-moment correlation coefficient $r$ .
For the pairs $(1, 2)$ , $(2, 3)$ , $(3, 5)$ , $(4, 4)$ and $(5, 6)$ , find the least-squares regression line $y = a + bx$ .
Work out $\dbinom{10}{3}$ .
The random variable $X \sim N(50, 5^2)$ . Work out the two $x$ -values of the points of inflection, which are at $\mu - \sigma$ and $\mu + \sigma$ .
The random variable $X \sim B(50, 0.4)$ . Use the Normal approximation, with the continuity correction, to estimate $P(X \le 15)$ .
The random variable $Z \sim N(0, 1)$ . Work out $P(-2 < Z < 2)$ .
The random variable $X \sim B(15, 0.8)$ . Use the Normal approximation to estimate $P(X \ge 12)$ .

Answers and concise solutions

1. The probability

Answer: The quantity is a count in a fixed 20 independent trials with constant $p = 0.8$ , so $X \sim B(20, 0.8)$ . Then $P(X \le 13) = 1 - P(X \ge 14) \approx 0.0867$ . So 13 or fewer seeds germinate in about 8.7% of packets.

2. The probability

Answer: The height is a measurement, so the Normal model fits. Standardise: $z = \dfrac{52 - 45}{6} = 1.1667$ . Then $P(X > 52) = 1 - \Phi(1.1667) \approx 0.1217$ . About 12.2% of plants are taller than 52 cm.

3. The probability

Answer: Add the terms for 0, 1 and 2. The exact value is $\tfrac{5888}{19683} \approx 0.2991$ . Keep four decimal places in each term and round once.

4. The probability

Answer: Standardise both ends: $z = \dfrac{62 - 70}{8} = -1$ and $z = \dfrac{78 - 70}{8} = 1$ . Then $P(62 < X < 78) = \Phi(1) - \Phi(-1) = 0.6827$ . The interval is the one-standard-deviation band.

5. The value of a

Answer: The value of $z$ with $\Phi(z) = 0.90$ is $z = 1.2816$ . Then $a = \mu + z\sigma = 100 + 1.2816 \times 15 = 119.224$ , so $a \approx 119.2$ (1 d.p.).

6. The correlation coefficient

Answer: The correlation coefficient is $r = \tfrac{9}{10} = 0.9$ . The points rise from left to right and lie close to a straight line, so the correlation is positive and strong. A strong correlation does not prove that one variable causes the other.

7. The line

Answer: The least-squares line is $y = 1.3 + 0.9x$ , so the gradient is $b = \tfrac{9}{10} = 0.9$ and the intercept is $a = \tfrac{13}{10} = 1.3$ . The gradient says that each extra unit of $x$ is associated with about 0.9 more of $y$ .

8. The value

Answer: $\dbinom{10}{3} = \dfrac{10!}{3!\,7!} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ . This coefficient appears in the binomial expansion and in the binomial probability formula.

9. The two values

Answer: The points of inflection are at $x = 45$ and $x = 55$ . At these points the curve is steepest and changes from concave to convex, so the standard deviation fixes where the curve turns.

10. The estimate

Answer: Here $np = 20$ and $n(1-p) = 30$ , both greater than 5, so the approximation is reasonable. Use $\mu = 20$ and $\sigma = \sqrt{12} \approx 3.464$ . The continuity correction turns $X \le 15$ into $Y \le 15.5$ , so $z = -1.30$ and $P(X \le 15) \approx 0.0970$ .

11. The probability

Answer: $P(-2 < Z < 2) = \Phi(2) - \Phi(-2) = 0.9772 - 0.0228 = 0.9545$ . About 95% of a Normal variable lies within two standard deviations of the mean.

12. The estimate

Answer: Here $np = 12$ and $n(1-p) = 3$ , so $n(1-p)$ is not greater than 5 and the approximation is weaker. The matching Normal variable has $\mu = 12$ and $\sigma = \sqrt{15 \times 0.8 \times 0.2} = \sqrt{2.4} \approx 1.549$ . The continuity correction turns $X \ge 12$ into $Y \ge 11.5$ , so $z \approx -0.323$ and the estimate is $0.6266$ . The exact binomial value is $0.6482$ , so the estimate is out by about 0.02. This shows why the conditions matter.

What you can now do

list the values and probabilities of a discrete random variable and check that they sum to 1;
test a context against the four binomial conditions and write $X \sim B(n, p)$ ;
work out a binomial probability for one value, an at-most range, an at-least range and a between range;
describe the Normal curve and its two parameters, and read the variance correctly in $N(\mu, \sigma^2)$ ;
standardise a value and read a probability from a standard Normal table;
find a value when you are given a probability;
explain how a histogram approaches a probability density curve, and why the points of inflection of a Normal curve sit at $\mu \pm \sigma$ ;
use the Normal approximation to the binomial with the continuity correction, and state when it is reasonable;
choose between the binomial and the Normal model for a context, and give a reason.

Your relevance reflection

Return to the prompt in the chapter opener: one situation where you would count how many times something happens in a fixed number of tries, and one where you would measure something that varies around an average.

Write both situations again. For each one, now add the model you would choose, the two parameters you would need, and one assumption you would have to check. Finish with one sentence about what you would do if the data did not match the model.