A certain train journey takes place every day throughout the year. The time taken, in minutes, for the journey is normally distributed with variance 11.2.
- (a)The mean time for a random sample of n of these journeys was found. A 94% confidence interval for the population mean time was calculated and was found to have a width of 1.4076 minutes, correct to 4 decimal places. Find the value of n.[3]
- (b)A passenger noted the times for 50 randomly chosen journeys in January, February and March. Give a reason why this sample is unsuitable for use in finding a confidence interval for the population mean time.[1]
- (c)A researcher took 4 random samples and a 94% confidence interval for the population mean was found from each sample. Find the probability that exactly 3 of these confidence intervals contain the true value of the population mean.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [M1] z * √(11.2/n) = 1.4076/2 (Any z, but must be a z.)
- [B1] z = 1.881 or 1.882
- [A1] n = 80 (Must be a whole number.)
- [B1] Jan, Feb and March not typical of whole year. (Or, e.g., weather is different at different times of year.)
- [M1] 0.94^3 * 0.06 * 4
- [A1] = 0.199 (3 sf)
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →