solution, P(X < 3) = P(0) + P(1) + P(2) M1
Binomial term with 11Cr p
= (0.84)11 + (0.16)(0.84)10 × 11C1 +
(0.16)2
(0.84)9
× 11C2
Correct expression for P(0, 1, 2) or P(0, 1, 2, 3)
= 0.1469 + 0.30782 + 0.2931
= 0.748
ii) μ = 125 × 0.64 = 80
σ^2= 125 × 0.64 × 0.36 = 28.8
P(X > 73) = 1 –phi(73.5-80/sqrt(28.8)
standardising = phi(1.211)
observe region, correct region (> 0.5 if mean > 73.5, if mean < 73.5)
= .887 ans
(i) Well, if we want to find the probability that fewer than 3 out of 11 vehicles are buses, we can start by finding the probability that exactly 0 buses are selected and the probability that exactly 1 bus is selected. Then, we can sum up those probabilities.
The probability of selecting exactly 0 buses out of 11 can be calculated using the binomial distribution formula: (11 choose 0) * (0.16)^0 * (0.84)^11.
The probability of selecting exactly 1 bus out of 11 can be calculated using the binomial distribution formula: (11 choose 1) * (0.16)^1 * (0.84)^10.
Finally, we can sum up these two probabilities to get the probability of selecting fewer than 3 buses.
(ii) Now, let's move on to the second part. We want to find the probability that more than 73 out of 125 vehicles are cars.
The problem becomes a bit trickier here because we're dealing with a large sample size. However, we can use a normal approximation to estimate the probability.
First, we find the mean and standard deviation of the number of cars using the given percentages. Then, we can use these values to calculate the probability of selecting more than 73 cars using the normal distribution.
However, I must warn you that my calculations might be as accurate as an elephant on roller skates. So, take these approximations with a pinch of humor!
