The distances travelled to work, D km, by the employees at a large company are normally distributed with $D \sim N(30, 8^2)$ - Edexcel - A-Level Maths Statistics - Question 7 - 2010 - Paper 2

To find the upper quartile $Q_3$, we can use the formula for quartiles in a normal distribution:

$Q_3 = \mu + Z_{0.75} \times \sigma$

Where $Z_{0.75} = 0.675$ (from standard normal distribution tables).

Thus,

$Q_3 = 30 + 0.675 \times 8 = 35.4.$

Thus, $Q_3 \approx 35.4$.

To find the lower quartile $Q_1$, we apply a similar method:

$Q_1 = \mu + Z_{0.25} \times \sigma$

Where $Z_{0.25} \approx -0.675$.

So,

$Q_1 = 30 - 0.675 \times 8 = 24.6.$

Thus, $Q_1 \approx 24.6$.

Using the quartiles calculated, we can find the outlier thresholds:

1. Calculate the interquartile range (IQR): $IQR = Q_{3} - Q_{1} = 35.4 - 24.6 = 10.8.$

2. Then, use it to find h and k:

   • For h: $h = Q_{1} - 1.5 \times IQR = 24.6 - 1.5 \times 10.8 = 24.6 - 16.2 = 8.4.$

   • For k: $k = Q_{3} + 1.5 \times IQR = 35.4 + 1.5 \times 10.8 = 35.4 + 16.2 = 51.6.$

Thus, $h \approx 8.4$ and $k \approx 51.6$.

An employee is selected at random. To find the probability that the distance travelled is an outlier, we consider values less than h or greater than k:

1. For $D < h$ (i.e., $D < 8.4$), we standardize: $Z = \frac{8.4 - 30}{8} = -2.675.$ Then, find the probability: $P(D < 8.4) = P(Z < -2.675) \approx 0.0037$ (using tables).

2. For $D > k$ (i.e., $D > 51.6$), we standardize: $Z = \frac{51.6 - 30}{8} = 2.675.$ Then find: $P(D > 51.6) = P(Z > 2.675) \approx 0.0037.$

3. Combine both probabilities for the total probability of being an outlier: $P(D < 8.4) + P(D > 51.6) = 0.0037 + 0.0037 = 0.0074.$

Thus, the probability that the distance travelled to work is an outlier is approximately $0.0074$.