Tiana is a quality controller in a clothes factory - AQA - A-Level Maths Mechanics - Question 18 - 2020 - Paper 3

For the number of shirts with a sewing defect, we have:

Let Y be the number of shirts with a sewing defect. Then,

$Y \sim B(30, 0.40)$

We need:

$P(Y < 15) = P(Y \leq 14)$

Using the binomial formula or a calculator, we find:

$P(Y < 15) \approx 0.8246$

To find the probability of at least 20 shirts without a fabric defect, we first find the probability for no fabric defect, which is:

The probability of not having a fabric defect is:

$1 - P(Fabric) = 1 - 0.30 = 0.70$

Let Z be the number of shirts without a fabric defect:

$Z \sim B(30, 0.70)$

We need:

$P(Z \geq 20) = 1 - P(Z \leq 19)$

Calculating gives us:

$P(Z \geq 20) \approx 0.2696$

For the hypothesis testing, we formulate:

Let X be the number of defective shirts with a fabric defect in a sample of size 60:

$X \sim B(60, 0.30)$

At the 5% level of significance, we will find the critical region:

$P(X \leq k) < 0.05$

Using binomial distribution tables or computational tools, we find:

The critical region is for x ≤ 11.

In her sample, Tiana finds 13 shirts with a fabric defect. Since 13 > 11, we do not reject H0:

Therefore, there is insufficient evidence to suggest that the proportion of shirts with a fabric defect has decreased.