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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

Answer :

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

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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