A customer service centre records every call they receive - AQA - A-Level Maths Pure - Question 14 - 2022 - Paper 3

To find $P(X = 1)$, we can use the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ Where:

• $n$ is the number of trials (20),
• $k$ is the number of successes (1), and
• $p$ is the probability of success (0.3).

Thus:

$P(X = 1) = \binom{20}{1} (0.3)^1 (0.7)^{19}$ Calculating this gives: $P(X = 1) = 20 \times 0.3 \times (0.7)^{19} \approx 0.00684.$

To find $P(X < 4)$, we can calculate: $P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).$ Calculating each:

• $P(X = 0) = \binom{20}{0} (0.3)^0 (0.7)^{20} \approx 0.000797.$
• $P(X = 1) = 0.00684$ (from previous calculation).
• $P(X = 2) = \binom{20}{2} (0.3)^2 (0.7)^{18} \approx 0.0367.$
• $P(X = 3) = \binom{20}{3} (0.3)^3 (0.7)^{17} \approx 0.1095.$

Summing these: $P(X < 4) \approx 0.000797 + 0.00684 + 0.0367 + 0.1095 \approx 0.1548.$

To calculate $P(X \geq 10)$, we can use: $P(X \geq 10) = 1 - P(X < 10).$

We would calculate $P(X < 10)$, which involves summing probabilities from $P(X = 0)$ to $P(X = 9)$. However, we can also use: $P(X < 10) = 1 - P(X \geq 10).$

And from the previous calculations for lower values, we can find: $P(X \geq 10) \approx 1 - 0.952 \approx 0.048.$

Given that the standard deviation of $Y$ is 1.5, we know:

$\sigma = \sqrt{np(1-p)}.$ Setting this equal to 1.5, we have:

$\sqrt{10p(1-p)} = 1.5.$ Squaring both sides: $10p(1-p) = 2.25.$ Rearranging this equation yields: $10p - 10p^2 = 2.25$ $10p^2 - 10p + 2.25 = 0.$

Using the quadratic formula: $p = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 10 \cdot -2.25}}{2 \cdot 10}.$

Calculating the discriminant and roots gives approximately: $p \approx 0.34 \text{ or } 0.66.$