Question

A random variable follows the continuous uniform distribution between 160 and 340 . Calculate the following quantities for the distribution. a) P(220≤x≤290) b) P(160≤x≤250) c) P(x>190) d) What are the mean and standard deviation of this distribution? a) P(220≤x≤290)= (Type an integer or decimal rounded to three decimal places as needed.) b) P(160≤x≤250)= (Type an integer or decimal rounded to three decimal places as needed.) c) P(x>190)= (Type an integer or decimal rounded to three decimal places as needed.) d) The mean of this distribution is (Type an integer or a decimal.) The standard deviation of this distribution is (Type an integer or decimal rounded to two decimal places as needed.)

Answer 1

a) P(220≤x≤290) = 0.389, b) P(160≤x≤250) = 0.500, c) P(x>190) = 0.833, d) Mean = 250, Standard Deviation ≈ 72.169

Step-by-step explanation:

a) P(220≤x≤290)

To find this probability, we need to find the proportion of the total range that falls between 220 and 290. The total range is 340 - 160 = 180. So, the probability is (290 - 220) / 180 = 70 / 180 = 0.389.

b) P(160≤x≤250)

Similarly, to find this probability, we need to find the proportion of the total range that falls between 160 and 250. So, the probability is (250 - 160) / 180 = 90 / 180 = 0.500.

c) P(x>190)

To find this probability, we need to find the proportion of the total range that falls above 190. So, the probability is (340 - 190) / 180 = 150 / 180 = 0.833.

d) Mean and Standard Deviation

The mean of a continuous uniform distribution is equal to the average of the minimum and maximum values, which is (160 + 340) / 2 = 250.
The standard deviation of a continuous uniform distribution is given by the formula sqrt((b - a)^2 / 12), where a is the minimum value and b is the maximum value. In this case, the standard deviation is sqrt((340 - 160)^2 / 12) = sqrt(82800 / 12) ≈ 72.169.