weer dat die nuwe presies 440 ml Volgens die dat die volume alverdeel is met ' \( n \) vyking 12. dat ' \( n \) ewekansige 430 ml koeldrank [2] dat die gemiddelde ekansige steekproef fll en 442 ml sal [3] [4] gebruik om ' n sekere ospitaal is 300 mense A toegedien, maar \( 0 \% \) vertrouensinterval mense, met hierdie A worker at Coocoo-Cola claims that the new "Buddy" Coocoo bottles don't hold exactly 440 ml of soft drink. According to the manufacturing process, he knows that the volume of soft drink in the bottles is normally distributed with mean 440 and standard deviation 12. 6.1 What is the probability that a randomly chosen bottle contains less than 430 ml of soft drink? [2] 6.2 What is the probability that, for a random sample of 49 bottles, the mean volume of soft drink is between 436 ml and 442 ml ? [3] Question 7 [4] Treatment A is used at hospitals to cure a certain disease. At one specific hospital, 300 people with this disease are given Treatment A, but only 267 of them are cured. Determine a \( 90 \% \) confidence interval for the population proportion of people, with this disease, who

Below is a step‐by‐step explanation and solution for each part of the problem. ───────────────────────────── 6.1. Probability that a Randomly Chosen Bottle Contains Less Than 430 ml Given:  Mean (µ) = 440 ml  Standard deviation (σ) = 12 ml We first standardize the value 430 ml using the z‐score formula:   z = (x – µ) / σ    = (430 – 440) / 12    = –10 / 12 ≈ –0.8333 Now, using the standard normal table (or a calculator) we look up the probability corresponding to z = –0.83. This probability is approximately 0.2023. Thus, the probability that a randomly chosen bottle contains less than 430 ml is about 0.202 (i.e. 20.2%). ───────────────────────────── 6.2. Probability that, for a Random Sample of 49 Bottles, the Mean Volume is Between 436 ml and 442 ml For a sample of size n = 49, the sampling distribution of the sample mean is normally distributed with:  Mean = 440 ml  Standard error (σₓ̄) = σ / √n = 12 / √49 = 12 / 7 ≈ 1.7143 ml Now, we standardize the lower and upper bounds. • For 436 ml:  z₁ = (436 – 440) / (12/7)    = (–4) / 1.7143 ≈ –2.3333 • For 442 ml:  z₂ = (442 – 440) / (12/7)    = 2 / 1.7143 ≈ 1.1667 Next, we find the probabilities:  P(Z < 1.17) is approximately 0.879  P(Z < –2.33) is approximately 0.0099 So, the probability that the sample mean is between 436 ml and 442 ml is:  P(436 < x̄ < 442) = P(–2.33 < Z < 1.17) ≈ 0.879 – 0.0099 ≈ 0.8691 Thus, the probability is about 0.869 (or 86.9%). ───────────────────────────── Question 7. 90% Confidence Interval for the Population Proportion (Treatment A) At the hospital, 300 people are treated and 267 are cured. The sample proportion is:   p̂ = 267 / 300 = 0.89 We want to construct a 90% confidence interval for the true population proportion, p. The confidence interval for a proportion is given by:   p̂ ± z* √[p̂(1 – p̂) / n] For a 90% confidence level, the z‑value (z*) is approximately 1.645. Now, calculate the standard error (SE):   SE = √[0.89 × (1 – 0.89) / 300]     = √[0.89 × 0.11 / 300]     = √[0.0979 / 300]     = √[0.00032633]     ≈ 0.01806 Now compute the margin of error (ME):   ME = z* × SE = 1.645 × 0.01806 ≈ 0.0297 Thus, the 90% confidence interval is:   Lower limit: 0.89 – 0.0297 ≈ 0.8603   Upper limit: 0.89 + 0.0297 ≈ 0.9197 So, the 90% confidence interval for the proportion of patients who are cured using Treatment A is approximately (0.860, 0.920). ───────────────────────────── Summary of Answers 6.1 Probability a bottle contains <430 ml: ≈ 0.202 (20.2%) 6.2 Probability that the sample mean (n = 49) is between 436 ml and 442 ml: ≈ 0.869 (86.9%) Question 7 The 90% confidence interval for the population proportion is approximately (0.860, 0.920).