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 obtained using the formula:

$Q₃ = \mu + Z_{0.75} \times \sigma$

Where Z_{0.75} is approximately 0.674 (for the 75th percentile). Substituting the values:

$Q₃ = 30 + 0.674 \times 8 = 30 + 5.392 = 35.4$

Thus, Q₃ is approximately 35.4.

The lower quartile Q₁ can be found similarly:

$Q₁ = \mu + Z_{0.25} \times \sigma$

Where Z_{0.25} is about -0.674. Therefore:

$Q₁ = 30 + (-0.674) \times 8 = 30 - 5.392 = 24.608$

So, Q₁ is approximately 24.6.

To find h and k, we use the formulas:

$h = Q₁ - 1.5 \times (Q₃ - Q₁)\n= 24.608 - 1.5 \times (35.4 - 24.608)\n= 24.608 - 1.5 \times 10.792\n= 24.608 - 16.188 = 8.42$

And for k:

$k = Q₃ + 1.5 \times (Q₃ - Q₁)\n= 35.4 + 1.5 \times (35.4 - 24.608)\n= 35.4 + 1.5 \times 10.792\n= 35.4 + 16.188 = 51.588$

Therefore, h is approximately 8.42 and k is approximately 51.59.

An outlier occurs if D < h or D > k. Therefore, we need to find:

$P(D < 8.42) + P(D > 51.59)$

For D < 8.42, we standardize:

$Z = \frac{8.42 - 30}{8} = -2.6775$

Looking up this value, we find:

$P(Z < -2.6775) \approx 0.003$

For D > 51.59:

$Z = \frac{51.59 - 30}{8} = 2.67375$

Looking this value up, we find:

$P(Z > 2.67375) \approx 0.003$

Finally, the total probability is:

$P(D < 8.42) + P(D > 51.59) \approx 0.003 + 0.003 = 0.006$

Thus, the probability that the distance travelled to work by the employee is an outlier is approximately 0.006.