Answer to Question #229026 in Statistics and Probability for nat

Question

3.2 Norman is a student at a college in Durban. The amount of time, in minutes, that Norman walks to the college for his final examinations is constantly distributed between 15 to 40 minutes, inclusive. Use this information to answer the following questions. 3.2.1 Name the continuous probability distribution described above. Explain in detail why it is called the distribution of little information. (3) 3.2.2 Calculate the probability that the student will take between 28 and 38 minutes. Provide interpretation for your answer. (2) 3.2.3 Find the probability that the student will take no more than 30 minutes to arrive at the college. Provide interpretation for your answer. (2) 3.2.4 Compute the probability that Norman will take least 35 minutes to get to the college. Provide interpretation for your answer. (2) 3.2.5 Calculate the mean, variance and the standard deviation of the distribution described in 3.2. 

Answer 1

3.2.1

This is a uniform distribution since then; the probability is constant between 15 and 40 minutes.

3.2.2

$P(28<x<38)\\ Probability = \frac{38-28}{40-15}=\frac{10}{25}=0.4$

The probability of the student taking between 28 and 38 minutes is 0.4

3.2.3

$P(x≤30)\\ Probability = \frac{30-15}{40-15}=\frac{15}{25}=0.6$

The probability of the student will take less than 30 minutes is 0.6

3.2.4

$P(x≥35)\\ Probability = \frac{40-35}{40-35}=\frac{5}{25}=0.2$

The probability of the student will take atleast 35 minutes is 0.2

3.2.5

Mean and Standard deviation

$Mean = \frac{a+b}{2}=\frac{40+15}{2}=27.5 \space minutes$

$Standard \space deviation = \sqrt{\frac{(b-a)^2}{12}}= \sqrt{\frac{(40-15)^2}{12}}=7.2169$

$Variance = \frac{(b-a)^2}{12}= \frac{(40-15)^2}{12}=\frac{625}{12}$