Amy has $30$ days of paid holiday, on which she decided to go for a bike journey in France. She chose a safe route of $2300$ kilometers.
Every morning Amy tosses a die, and if the outcome is $k$, where $k = \{1,2,3,4,5,6\}$, she cycles $20k$ kilometers that day.
In which of the following intervals falls the probability that Amy will complete her route within $30$ days (or less): [6 marks]
i) $[0, 0.25]$
ii) $[0.25, 0.5]$
iii) $[0.5, 0.75]$
iv) $[0.75, 1]$
Late edit but I thought I'd update my final solution here and interpretation to the intervals $[0, 0.25]$ since I believe the answers are misinterpreting the question e.g. $P[\bar X<30]$, but not in the interval 22.5-30 days ($P[22.5<\bar X<30]$. I am not sure if I have interpreted these intervals correctly as well, or if they are meant to be standard deviations or something.
Solution:
- Calculating the expected distance travel per day: $\mu = 1/6(20+40+60+80+100+120)=70km$
- Translating this to the expected days required (not sure if this is correctly done, but I did): $2300/70 = 32.86$ days
Calculating the standard deviation to be $34.1$ $\sigma = \sqrt{\frac{(20-70)^2+(40-70)^2+(60-70)^2+(80-70)^2+(100-70)^2+(120 -70)^2}6} = 34.1$
Then by applying the central limit theorem, because as n reaches values greater than 30, $\bar x $ tends to the normal distribution $(\mu, \frac{\sigma}{\sqrt n})$ This also means new standard deviation of the sample mean, $\bar x$ is $6.24$
By applying the formulae $Z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}$, and drawing the normalised curve for each interval, I attempted to calculate the area between the 4 intervals using the Z - table.
The only thing is this required a lot of work for a mere 6 marks and I am not sure if I have tackled it in the required way! This is an example question for a first year undergrad.
The mean number of km she does is $3.5 \cdot 20 \cdot 30=2100=\mu$ The variance on one day is $\frac{35}{12}$, so the total variance is $\frac {35 \cdot 30}{12}=\frac {175}2$. The standard deviation is $\sigma=\sqrt{\frac {175}2}$ She needs to be $\frac {2300-2100}{\sigma}\approx 21.38$ standard deviations high. Ain't gonna happen