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

Work through these on your own. Check $np > 5$ and $n(1-p) > 5$ before you approximate, and use the continuity correction. For a general Normal variable, $\mu - \sigma$ and $\mu + \sigma$ are the points of inflection.

The questions and the answers below come from a set that is checked by computer every time the book is built.

The random variable $X \sim B(50, 0.4)$ . Work out the mean $\mu = np$ of the matching Normal variable.
The random variable $X \sim B(50, 0.4)$ . Work out the variance $np(1-p)$ of the matching Normal variable.
The random variable $X \sim B(50, 0.4)$ . Work out the standard deviation $\sigma = \sqrt{np(1-p)}$ of the matching Normal variable.
The random variable $X \sim B(50, 0.4)$ . Use the Normal approximation to estimate $P(X \le 15)$ .
The random variable $X \sim B(100, 0.5)$ . Work out the mean of the matching Normal variable.
The random variable $X \sim B(100, 0.5)$ . Work out the standard deviation of the matching Normal variable.
The random variable $X \sim B(100, 0.5)$ . Use the Normal approximation to estimate $P(X \le 40)$ .
The random variable $X \sim N(\mu, \sigma^2)$ . Work out $P(\mu - \sigma < X < \mu + \sigma)$ , the probability within one standard deviation of the mean.
The random variable $X \sim N(\mu, \sigma^2)$ . Work out $P(\mu - 2\sigma < X < \mu + 2\sigma)$ .
The random variable $X \sim N(\mu, \sigma^2)$ . Work out $P(\mu - 3\sigma < X < \mu + 3\sigma)$ .
The random variable $X \sim N(50, 5^2)$ . The Normal curve has points of inflection at $\mu - \sigma$ and $\mu + \sigma$ . Work out the two values of $x$ .
The random variable $X \sim B(50, 0.4)$ . Use the Normal approximation to estimate $P(X \le 24)$ .
The random variable $X \sim B(50, 0.4)$ . Work out the exact binomial probability $P(X \le 15)$ , to compare with the Normal estimate.

Answers and concise solutions

1. The mean

Answer: $\mu = np = 50 \times 0.4 = 20$ . The centre of the Normal curve sits at the centre of the binomial bars.

2. The variance

Answer: $np(1-p) = 50 \times 0.4 \times 0.6 = 12$ . The variance is 12, so the standard deviation is $\sqrt{12}$ .

3. The standard deviation

Answer: $\sigma = \sqrt{12} = 2\sqrt{3} \approx 3.464$ . Both $np = 20$ and $n(1-p) = 30$ are greater than 5, so the approximation is reasonable.

4. The estimate

Answer: Apply the continuity correction: $X \le 15$ becomes $Y \le 15.5$ . Then $z = \dfrac{15.5 - 20}{3.464} \approx -1.30$ , and $P(X \le 15) \approx \Phi(-1.30) = 0.0968$ . To four decimal places the estimate is 0.0970. The exact binomial value is 0.0955.

5. The mean

Answer: $\mu = np = 100 \times 0.5 = 50$ . With $p = 0.5$ the distribution is symmetric, so the curve is centred on 50.

6. The standard deviation

Answer: $\sigma = \sqrt{np(1-p)} = \sqrt{100 \times 0.5 \times 0.5} = \sqrt{25} = 5$ . Both $np$ and $n(1-p)$ equal 50, so the approximation conditions hold.

7. The estimate

Answer: Apply the continuity correction: $X \le 40$ becomes $Y \le 40.5$ . Then $z = \dfrac{40.5 - 50}{5} = -1.9$ , and $P(X \le 40) \approx \Phi(-1.9) = 1 - 0.9713 = 0.0287$ .

8. The probability

Answer: $P(\mu - \sigma < X < \mu + \sigma) = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6827$ , so about 68% of a Normal variable lies within one standard deviation of the mean.

9. The probability

Answer: $P(\mu - 2\sigma < X < \mu + 2\sigma) = \Phi(2) - \Phi(-2) = 0.9772 - 0.0228 = 0.9545$ , so about 95% lies within two standard deviations.

10. The probability

Answer: $P(\mu - 3\sigma < X < \mu + 3\sigma) = \Phi(3) - \Phi(-3) = 0.9987 - 0.0013 = 0.9973$ , so about 99.7% lies within three standard deviations.

11. The two values of x

Answer: The points of inflection are at $x = \mu - \sigma = 50 - 5 = 45$ and $x = \mu + \sigma = 50 + 5 = 55$ . These are the points where the curve is steepest and changes from concave to convex.

12. The estimate

Answer: Apply the continuity correction: $X \le 24$ becomes $Y \le 24.5$ . Then $z = \dfrac{24.5 - 20}{3.464} \approx 1.30$ , and $P(X \le 24) \approx \Phi(1.30) = 0.9030$ .

13. The exact probability

Answer: The exact binomial probability is $P(X \le 15) = 0.0955$ (4 d.p.). The Normal estimate of 0.0970 is close, which is why the approximation is useful when $n$ is large.