Question:
"An aeroplane is licensed to carry 100 passengers. If the weights of passengers are normally distributed with a mean of 80 kg and a standard deviation of 20 kg.Find the probability that the combined weight of 100 passengers will exceed 8500.
What I have attempted:
µ = 85, σ = 20, n = 100
Pr(100X̄>8500)
Pr(X̄>85)
And by dividing σ by root(100), σ of X̄ is shown to be 20/10, from this the standardized distribution Z is gained.
Pr(Z>5/2), this equals 0.0062, or 0.62%
The answer supposedly is 0.0138, or 1.38%.
Obviously this doesn't work. So I tried something else.
I looked into confidence interals:
We know that µ+k(σ/root(n)) = the upper terminal of the confidence interval.
If we sub in what we know we can theoretically work out what the k value is and work out the probability of being greater than this k value on a standard distribution.
80+k(20/10)=85
80+k(2)=85
k=5/2
We obtain the same result. Can someone please show me the error in my thinking?
Looks good to me: $$ \Bbb{P}\Big(\frac{\sum_{i=1}^{100} X_i - 8000}{ 10 \times 20 } > \frac{8500 - 8000}{ 10 \times 20 } \Big) = \Bbb{P}\Big( Z > \frac{500}{200}\Big) = \Bbb{P}\Big( Z > \frac{5}{2}\Big) $$