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A machine puts liquid into bottles of perfume - Edexcel - A-Level Maths Statistics - Question 5 - 2019 - Paper 1
Question 5
A machine puts liquid into bottles of perfume. The amount of liquid put into each bottle, Dml, follows a normal distribution with mean 25 ml Given that 15% of bottl... show full transcript
Worked Solution & Example Answer:A machine puts liquid into bottles of perfume - Edexcel - A-Level Maths Statistics - Question 5 - 2019 - Paper 1
Step 1
(a) find, to 2 decimal places, the value of k such that P(24.63 < D < k) = 0.45
Answer
To solve for k, we first determine the standard deviation (σ) of the distribution. Given that 15% of bottles contain less than 24.63 ml, we look for the z-score that corresponds to the 15th percentile:
z = rac{x - ext{mean}}{ ext{std dev}}
Given: x = 24.63, mean = 25; We find the z-score using a z-table:
ightarrow z ext{ for } 0.15 = -1.0364$$ Now we can set it up: $$-1.0364 = rac{24.63 - 25}{ ext{std dev}}$$ Solving gives: $$ ext{std dev} ext{ (σ)} = 0.357$$ Now we want: $$P(24.63 < D < k) = 0.45$$ So, this means: $$P(D < k) = P(D < 24.63) + 0.45 = 0.15 + 0.45 = 0.60$$ The z-score for 0.60 is approximately 0.2533: $$k = ext{mean} + (z * ext{std dev}) ightarrow k = 25 + 0.2533(0.357) ightarrow k ext{ (to 2 d.p.)} ext{ is approximately } 25.09$$Step 2
(b) Using a normal approximation, find the probability that fewer than half of these bottles contain between 24.63 ml and k ml
Answer
In this situation, we want to determine whether fewer than 100 bottles (i.e., half of 200) contain liquid between 24.63 ml and k ml.
Using the normal approximation: We have:
- Mean (μ) = 200 × P(24.63 < D < k) = 200 × 0.45 = 90
- Standard deviation (σ) is computed as follows:
Step 3
(c) Test Hannah’s belief at the 5% level of significance. You should state your hypotheses clearly.
Answer
Here, we set the null and alternative hypotheses:
- Null Hypothesis (H0): μ = 25 ml
- Alternative Hypothesis (H1): μ < 25 ml
We will calculate the z-score using Hannah's sample mean: z = rac{ar{x} - ext{mean}}{ rac{ ext{std dev}}{ ext{sqrt}(n)}} Given:
- Sample mean (x̄) = 24.94 ml
- Standard deviation (σ) = 0.16 ml
- n = 20 bottles
So, z = rac{24.94 - 25}{ rac{0.16}{ ext{sqrt}(20)}} Calculating this gives:
Referring to the z-table, the critical z-value for a one-tailed test at 5% significance is approximately -1.645. Since -0.6466 > -1.645, we fail to reject the null hypothesis. Therefore, there is insufficient evidence to support Hannah’s belief that the mean amount of liquid put in each bottle is less than 25 ml.