A health centre claims that the time a doctor spends with a patient can be modelled by a normal distribution with a mean of 10 minutes and a standard deviation of 4 minutes - Edexcel - A-Level Maths Statistics - Question 5 - 2020 - Paper 1

We set up the following hypotheses:

• Null Hypothesis (H₀): μ = 10
• Alternative Hypothesis (H₁): μ > 10

We calculate the test statistic using the sample mean ar{x} = 11.5 minutes, sample size n = 20, and the known population standard deviation σ = 4:

$Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{11.5 - 10}{\frac{4}{\sqrt{20}}}$

Calculating the values:

$Z \approx \frac{1.5}{\frac{4}{4.472}} = \frac{1.5}{0.8944} \approx 1.676.$

Using a Z-table for a one-tailed test at α = 0.05, the critical value is approximately 1.645. Since 1.676 > 1.645, we reject the null hypothesis. Thus, at the 5% significance level, there is evidence to support the patients’ complaint.

We know T follows the distribution T ~ N(5, 3.5²). To find the probability that a routine appointment with the dentist takes less than 2 minutes:

$P(T < 2) = P\left(Z < \frac{2 - 5}{3.5}\right)$.

Calculating Z:

$Z = \frac{2 - 5}{3.5} \approx -0.857.$

Using a Z-table, we have:

$P(Z < -0.857) \approx 0.195.$

Therefore, the probability that the appointment takes less than 2 minutes is approximately 0.195.

To find this conditional probability, we can use the following formula:

$P(T < 2 | T > 0) = \frac{P(T < 2) \cdot P(T > 0)}{P(T > 0)}$.

From the previous part, we found:

$P(T < 2) \approx 0.195$.

Next, we need to calculate P(T > 0):

Using the Z-score:

$P(T > 0) = P\left(Z > \frac{0 - 5}{3.5}\right) = P(Z > -1.43).$

From the Z-table:

$P(Z > -1.43) \approx 0.923.$

Combining these:

$P(T < 2 | T > 0) = \frac{0.195 \cdot 0.923}{0.923} = 0.195.$

So, $P(T < 2 | T > 0) \\approx 0.195$.

The normal distribution may not be a good model for T if we observe that a significant portion of values fall below 0. Since the time for a routine appointment cannot realistically be negative, using a normal distribution can produce probabilities that suggest negative times are feasible, which is nonsensical. The model’s assumptions fail since it does not appropriately represent the constraints of the situation.

For the refined model considering only T > 2, the median of a truncated normal distribution can be found by finding the z-value which corresponds to the 0.5 probability limit:

Using the formula for the median:

$M = \mu + \sigma Z$

Where Z is determined by the truncated limits. However, we want the median from a cutoff point, hence we would need to find where:

• For values greater than 2, we take the mean which effectively shifts the median upwards. Since the exact statistical calculation may involve more complex statistical tables, with the approximation:

Given μ = 5 and σ = 3.5, The median is approximately around 5.9 for the new model rounding to one decimal place. Thus, the median time for a routine appointment with the dentist is 5.9 minutes.