A study of the time spent shopping in a supermarket for a market basket of 20 specific items showed an approximately uniform distribution between 20 minutes and 40 minutes. What is the probability that the shopping rime will be

a. Between 25 and 30 minutes?
b. less than 35 minutes?
c. What are the mean and standard deviation of the shopping time?

To find the probability in a uniform distribution, we need to calculate the area under the probability density function (PDF) curve between the given intervals.

a. To find the probability that the shopping time is between 25 and 30 minutes, we need to calculate the area under the PDF curve between these two values. In a uniform distribution, the PDF is a constant value within the given interval and zero outside that interval.

The interval for the distribution is from 20 to 40 minutes, and it is uniform, so the height of the PDF curve is 1/(40-20) = 1/20. The width of the interval between 25 and 30 minutes is 5 minutes.

The probability between 25 and 30 minutes is equal to the area of a rectangle with height 1/20 and width 5. Therefore, the probability is (1/20) * 5 = 1/4 = 0.25.

b. To find the probability that the shopping time is less than 35 minutes, we need to calculate the area under the PDF curve up to 35 minutes. Since the distribution is uniform, the height of the PDF curve is 1/20, and the width of the interval is 35 - 20 = 15 minutes.

The probability up to 35 minutes is equal to the area of a rectangle with height 1/20 and width 15. Therefore, the probability is (1/20) * 15 = 3/4 = 0.75.

c. In a uniform distribution, the mean (μ) is the average of the lower and upper bounds of the interval of the distribution. In this case, the mean of the shopping time is (20 + 40)/2 = 30 minutes.

The standard deviation (σ) of a uniform distribution can be calculated using the formula: σ = (√3) * (upper bound - lower bound)/2. In this case, the standard deviation is (√3) * (40 - 20)/2 = 5√3 = 8.66025 minutes.

Therefore, the mean shopping time is 30 minutes, and the standard deviation is approximately 8.66025 minutes.