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------------------------------------------------ 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 E(X)=188 and V(X)=141 None of these E(X)= 141 and V(X)= 188 E(X)=564 and V(x)=141

To calculate the expected value (mean) and variance of the number of speeding cars passing the speed camera, we use the properties of the binomial distribution. The probability of a car speeding is given as 0.25, and the total number of cars passing the camera is 752.

The expected value (E(X)) of a binomial distribution is equal to n * p, where n is the number of trials (752 cars) and p is the probability of success (0.25). Therefore, E(X) = 752 * 0.25 = 188.

The variance (V(X)) of a binomial distribution is equal to n * p * (1 - p). Therefore, V(X) = 752 * 0.25 * (1 - 0.25) = 141.

So, you would expect 188 cars to be speeding, and the variance would be 141 for the 752 cars passing the speed camera.

Learn more about Variance

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