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
find, to 2 decimal places, the value of k such that $P(24.63 < D < k) = 0.45$
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Answer
To find the value of k, we first standardize the variable. Given that the mean, (\mu = 25) ml and the corresponding value for 24.63 ml:
Standardize 24.63 ml:
z=σ24.63−25=−1.0364
From the standard normal distribution, we know that 15% of bottles contain less than 24.63 ml. Therefore, we set:
P(Z<−1.0364)=0.15
We want to find k such that:
P(24.63<D<k)=0.45
This can be expressed as:
P(D<k)=P(D<24.63)+0.45=0.15+0.45=0.60
Now, we find from the standard normal table:
P(Z<zk)=0.60⇒zk=0.2533
Reversing the z-score calculation:
k=μ+zk⋅σ=25+0.2533⋅σ
Using (\sigma = 0.357):
k=25+0.2533⋅σ≈25.09
Therefore, the value of k is approximately 25.09 ml.
Step 2
Using a normal approximation, find the probability that fewer than half of these bottles contain between 24.63 ml and k ml
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Answer
Given the sample size of 200 bottles, we define a new variable Y:
Define the distribution: (Y \sim B(200, 0.45)) which will be approximated as (Y \sim N(90, 49.5)) since:
Probability that fewer than half (100 bottles) contain between 24.63 ml and k ml:
(\Rightarrow P(Y < 100) = P\left(Z < \frac{100 - 90}{\sqrt{49.5}}\right))
Calculating the z-score:
z=49.5100−90≈1.4142
Using the standard normal distribution table:
Find the probability:
P(Z<1.4142)≈0.9115(or 0.912 when approximated)
Step 3
Test Hannah's belief at the 5% level of significance. You should state your hypotheses clearly.
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Answer
State the null hypothesis and alternative hypothesis:
Null Hypothesis: (H_0: \mu \geq 25) (Hannah's belief is not supported)
Alternative Hypothesis: (H_1: \mu < 25) (Hannah's belief is supported)
Given the sample mean (\bar{x} = 24.94) and sample size (n = 20):
Standard deviation (s = 0.16)
Calculate the test statistic:
t=s/nxˉ−μ0=0.16/2024.94−25≈−1.614
Determine the critical value from the t-distribution table for (df = 19) at the 5% significance level (one-tailed test):
Critical value approximately -1.729.
Conclusion: Since (-1.614 > -1.729), we fail to reject the null hypothesis. There is insufficient evidence to support Hannah's belief.