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

Let:

• $\mu_0 = 10$ (hypothesized mean)
• $\bar{x} = 11.5$ (sample mean)
• $n = 20$ (sample size)
• $\sigma = 4$ (standard deviation)

The null hypothesis is:

• $H_0$: $\mu \leq 10$

The alternative hypothesis is:

• $H_a$: $\mu > 10$

Calculating the test statistic:

$Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{11.5 - 10}{4 / \sqrt{20}} \approx 1.77$

Finding the critical value at a 5% significance level (one-tailed test):

• Critical Z-value is approximately 1.645.

Since $Z = 1.77 > 1.645$, we reject $H_0$.

Conclusion: There is enough evidence at the 5% significance level to support the patients’ complaint.

To find this probability, we calculate the Z-score:

$Z = \frac{T - \mu}{\sigma} = \frac{2 - 5}{3.5} \approx -0.8571$

Using the standard normal distribution, we find:

$P(T < 2) = P(Z < -0.8571) \approx 0.1952$

Thus, the probability that a routine appointment takes less than 2 minutes is approximately 0.195.

Using the formula for conditional probability:

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

Calculating each part:

1. For $P(T < 2 \cap T > 0)$, we have $P(0 < T < 2) \approx P(Z < -0.8571) \approx 0.1952$
2. For $P(T > 0)$:
   • This can be computed as: $P(Z > -1.4286) \approx 1 - P(Z < -1.4286) \approx 0.9234$

Now substituting back:

$P(T < 2 | T > 0) = \frac{0.1952}{0.9234} \approx 0.211$

The computed probabilities indicate that while we find a small probability for T to be under 2 minutes, there remains a significant probability of T being greater than 0. However, with values like T < 2 having a not negligible probability, it suggests unrealistic scenarios within the context of dental appointments, as appointments usually require more time.

Given that we are interested only in values where T > 2, we need to find the median of the truncated distribution. The median for a normal distribution is equal to the mean for symmetric cases. The adjusted mean considering only T > 2 needs to account for the probability density until 2.

Using numerical methods or simulations, we can derive this median, leading to a computed value:

$\text{Median} \approx 5.9$

Thus, the median time for a routine appointment using this new model is approximately 5.9 minutes.