High School

The probability that cars passing a speed camera are speeding is 0.25. If 752 cars pass the camera, how many of the cars would you expect to be speeding and what would be the Variance? O None of these O E(X)=188 and V(x)=141 O E(X)=564 and V(x)=141 O E(X)= 141 and V(X)= 188

Answer :

You would expect 188 cars to be speeding, and the Variance would be 141.

1. Expected number of speeding cars (E(X)) is calculated by multiplying the total number of cars (752) by the probability of cars speeding (0.25): E(X) = 752 * 0.25 = 188 cars.

2. Variance (V(X)) is a measure of the spread of the data from the mean. It is calculated by multiplying the probability of success (p) with the probability of failure (q) and then multiplying the result by the total number of trials (n): V(X) = p * q * n.

3. In this case, p is the probability of cars speeding (0.25), and q is the probability of cars not speeding (1 - 0.25 = 0.75). Therefore, V(X) = 0.25 * 0.75 * 752 = 141.

In summary, you would expect 188 cars to be speeding based on the given probability and the number of cars passing the speed camera. The variance of the data is 141, indicating the spread of the number of speeding cars from the expected value.

Learn more about Variance.

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