A sign on the petrol pumps of a chain of petrol stations encourages customers to have their oil checked, claiming that one out of every four cars should have its oil topped up. Assuming this claim is true:
(b) What is the probability that at least half of the next 10 cars entering a petrol station should have their oil topped up? At least half of the next 20 cars?
(c) What is the probability that at least 5 but not more than 8 of the next 20 cars will need their oil topped up?
(d) What is the mean and standard deviation of the number of cars out of the next 20 that will need their oil topped up?
To solve these probability problems, we can use the binomial distribution. The binomial distribution is used when we have a fixed number of trials (in this case, the number of cars entering the petrol station) and each trial (car) has only two possible outcomes (either needing the oil topped up or not).
Let's start with part (b):
We want to find the probability that at least half (i.e., 5 or more) of the next 10 cars entering the petrol station should have their oil topped up.
To solve this, we can use the binomial probability formula:
P(X ≥ k) = 1 - P(X < k)
where X is the random variable representing the number of cars needing oil top-up, and k is the desired number of cars.
The probability of a car needing oil top-up is 1/4, as given in the problem.
For 10 cars, we need to calculate:
P(X ≥ 5) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4))
To calculate each term, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where (n choose k) is the binomial coefficient, given by n! / (k! * (n-k)!), p is the probability of success (need oil top-up), and n is the number of trials (cars).
Using this formula, we can calculate each term and sum them up. The same procedure can be followed for part (c) and part (d).
For part (c):
We want to find the probability that at least 5 but not more than 8 of the next 20 cars will need their oil topped up.
Here, we need to calculate:
P(5 ≤ X ≤ 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
For part (d):
We need to find the mean and standard deviation of the number of cars out of the next 20 that will need their oil topped up.
The mean (expected value) can be calculated using the formula:
μ = n * p
where μ is the mean, n is the number of trials, and p is the probability of success.
The standard deviation can be calculated using the formula:
σ = sqrt(n * p * (1-p))
where σ is the standard deviation.
By substituting the values n = 20 and p = 1/4 into these formulas, we can find the mean and standard deviation.
Remember to calculate probabilities, mean, and standard deviation, you will need to substitute the appropriate values into the formulas and perform the necessary calculations.