13.4 Links to the binomial, histograms and points of inflection — Practise with help

Work each question closed book first. Check the conditions $np > 5$ and $n(1-p) > 5$ before you approximate. Apply the continuity correction before you standardise.

Warm-up: remember

The random variable $Z \sim N(0, 1)$ . Find $P(Z < 1)$ .
A data set has mean 40 and standard deviation 5. How many standard deviations above the mean is the value 47?
Find $f''(x)$ for $f(x) = x^3 - 3x^2$ , and state the value of $x$ where $f''(x) = 0$ .

Answers

From the table, $\Phi(1) = 0.8413$ .
$\dfrac{47 - 40}{5} = 1.4$ standard deviations.
$f'(x) = 3x^2 - 6x$ , so $f''(x) = 6x - 6$ . Then $6x - 6 = 0$ gives $x = 1$ .

If question 3 was hard, revise the second derivative from Section 7.4.

Practise with help

1. The random variable $X \sim B(20, 0.5)$ . Find the mean and the standard deviation of the matching Normal variable.

Hint

Use $\mu = np$ and $\sigma = \sqrt{np(1-p)}$ .

Next step

Here $n = 20$ and $p = 0.5$ , so $np = 10$ and $n(1-p) = 10$ . Substitute into the formula for $\sigma$ .

Full working

\mu = np = 20 \times 0.5 = 10, \qquad \sigma = \sqrt{np(1-p)} = \sqrt{20 \times 0.5 \times 0.5} = \sqrt{5} \approx 2.236.

Both $np$ and $n(1-p)$ equal 10, so the approximation conditions hold.

2. Use the matching Normal variable from question 1 to estimate $P(X \le 7)$ for $X \sim B(20, 0.5)$ .

Hint

Apply the continuity correction first: $X \le 7$ becomes $Y \le 7.5$ .

Next step

Standardise with $\mu = 10$ and $\sigma = \sqrt5$ : $z = \dfrac{7.5 - 10}{\sqrt5} \approx -1.118$ .

Full working

z = \frac{7.5 - 10}{\sqrt5} \approx -1.118.

Then

P(X \le 7) \approx \Phi(-1.118) \approx 0.1318.

The exact binomial probability is $0.1316$ , so the estimate is close.

3. Use the same Normal variable to estimate $P(X \ge 13)$ for $X \sim B(20, 0.5)$ .

Hint

Apply the continuity correction: $X \ge 13$ becomes $Y \ge 12.5$ .

Next step

Standardise: $z = \dfrac{12.5 - 10}{\sqrt5} \approx 1.118$ . Then use the complement.

Full working

z = \frac{12.5 - 10}{\sqrt5} \approx 1.118.

P(X \ge 13) \approx 1 - \Phi(1.118) \approx 1 - 0.8682 = 0.1318.

The answer equals the answer to question 2. The distribution is symmetric because $p = 0.5$ , and the two boundaries 7.5 and 12.5 are the same distance from the mean 10.

4. The random variable $X \sim B(100, 0.3)$ . Find the mean and the standard deviation of the matching Normal variable.

Hint

Use $\mu = np$ and $\sigma = \sqrt{np(1-p)}$ , with $n = 100$ and $p = 0.3$ .

Next step

$np = 30$ and $n(1-p) = 70$ . Substitute into the formula for $\sigma$ .

Full working

\mu = np = 100 \times 0.3 = 30, \qquad \sigma = \sqrt{100 \times 0.3 \times 0.7} = \sqrt{21} \approx 4.583.

Both $np = 30$ and $n(1-p) = 70$ are greater than 5, so the approximation is reasonable.

5. Use the matching Normal variable to estimate $P(X \le 25)$ for $X \sim B(100, 0.3)$ .

Hint

Apply the continuity correction: $X \le 25$ becomes $Y \le 25.5$ .

Next step

Standardise: $z = \dfrac{25.5 - 30}{4.583} \approx -0.982$ . Then read the table.

Full working

z = \frac{25.5 - 30}{4.583} \approx -0.982.

P(X \le 25) \approx \Phi(-0.982) \approx 0.1631.

The exact binomial probability is $0.1631$ to four decimal places, so the approximation is very close.

6. The random variable $X \sim B(10, 0.1)$ . Explain why a Normal approximation is not appropriate here.

Hint

Check the two conditions $np > 5$ and $n(1-p) > 5$ .

Next step

Work out $np$ and $n(1-p)$ and compare each with 5.

Full working

np = 10 \times 0.1 = 1, \qquad n(1-p) = 10 \times 0.9 = 9.

The value $np = 1$ is not greater than 5. The distribution is strongly skewed to the right: the bar for 0 is much taller than the bar for 1, which is taller than the bar for 2. A symmetric Normal curve fits these bars badly. Use the binomial formula instead.