A machine puts liquid into bottles of perfume - Edexcel - A-Level Maths: Statistics - Question 5 - 2019 - Paper 1

In this part, we need to find the probability that fewer than 100 out of 200 bottles (half) contain between 24.63 ml and k ml (which we've calculated as 25.09 ml).

This can be approximated using the normal distribution:

Let ( X \sim B(200, 0.6) )

The mean (( \mu )) and variance (( \sigma^2 )) for this binomial distribution are:

• ( \mu = np = 200 \times 0.6 = 120 )
• ( \sigma^2 = np(1-p) = 200 \times 0.6 \times 0.4 = 48 )

Thus, the standard deviation (( \sigma )) is:

• ( \sigma = \sqrt{48} \approx 6.93 )

Now we calculate the z-score for 100:

$z = \frac{100 - 120}{6.93} \approx -2.89$

Using z-tables, we find:

• ( P(Z < -2.89) \approx 0.0019 )

Thus, the probability that fewer than half of these bottles contain between 24.63 ml and 25.09 ml is approximately 0.0019.

We will perform a hypothesis test concerning the mean liquid put in bottles following the adjustments:

Hypotheses:

• Null Hypothesis (H0): ( \mu \geq 25 ) ml (Hannah's belief is incorrect)
• Alternative Hypothesis (H1): ( \mu < 25 ) ml (Hannah's belief is correct)

We calculate the z-score using the sample mean (( \bar{x} = 24.94 )), sample size (n = 20), and standard deviation (( \sigma = 0.16 )):

$z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{24.94 - 25}{0.16 / \sqrt{20}}$

Calculating:

• $z = \frac{-0.06}{0.0358} \approx -1.678$

Using z-tables, we check the p-value for z = -1.678, which corresponds to approximately 0.04676.

Since this p-value (0.04676) is less than the significance level of 0.05, we reject the null hypothesis.

Conclusion: There is sufficient evidence to support Hannah's belief that the mean amount of liquid in each bottle is less than 25 ml.