The distances travelled to work, D km, by the employees at a large company are normally distributed with D ~ N(30, 8²) - Edexcel - A-Level Maths Statistics - Question 7 - 2010 - Paper 2

The upper quartile, Q3, is the value below which 75% of the data fall. To find Q3, we first find the Z-score for the 75th percentile:

From Z-tables or using a standard normal calculator, we find:

$Z_{75} = 0.675$

Using the formula for inverse transformation: $Q3 = \mu + Z * \sigma$ Substituting the values: $Q3 = 30 + 0.675 * 8 = 30 + 5.4 = 35.4$

Therefore, Q3 is approximately 35.4 km.

The lower quartile, Q1, is the value below which 25% of the data fall. Finding the Z-score for the 25th percentile:

From Z-tables, we have: $Z_{25} = -0.675$

Using the inverse transformation: $Q1 = \mu + Z * \sigma$ Substituting the values: $Q1 = 30 + (-0.675) * 8 = 30 - 5.4 = 24.6$

Thus, Q1 is approximately 24.6 km.

To find h and k, we will use the formulas:

1. Calculate Q3 - Q1: $Q3 - Q1 = 35.4 - 24.6 = 10.8$
2. Calculate h: $h = Q1 - 1.5 × (Q3 - Q1) = 24.6 - 1.5 × 10.8 = 24.6 - 16.2 = 8.4$
3. Calculate k: $k = Q3 + 1.5 × (Q3 - Q1) = 35.4 + 1.5 × 10.8 = 35.4 + 16.2 = 51.6$

Thus, $h$ is 8.4 and $k$ is 51.6.

An outlier is defined as a distance $D$ such that $D < h$ or $D > k$. We calculated:

• $h = 8.4$
• $k = 51.6$.

Now, we need to find: $P(D < 8.4) + P(D > 51.6)$ Using standardization:

1. For $D < 8.4$: $Z_h = \frac{8.4 - 30}{8} = -2.675$ So, $P(Z < -2.675) \approx 0.0037$.
2. For $D > 51.6$: $Z_k = \frac{51.6 - 30}{8} = 2.675$ Thus, $P(Z > 2.675) \approx 0.0037$ (symmetry of normal distribution).

Now, adding both probabilities: $P(D < 8.4) + P(D > 51.6) = 0.0037 + 0.0037 = 0.007$.

Thus, the probability that the distance travelled to work by this employee is an outlier is approximately 0.007.