The time (in minutes) until the next bus departs a major bus depot follows a uniform distribution from 25 to 46 minutes. Let X denote the time until the next bus departs. The distribution is and is . The mean of the distribution is μ= . The standard deviation of the distribution is σ= . The probability that the time until the next bus departs is between 30 and 40 minutes is P(30

Step-by-step explanation:

The distribution is Uniform and is continuous.

We are given a = 25 and b = 446

The mean of the distribution is

Mean = µ = (a + b) / 2

= (25 + 44) / 2

= 34.5

The standard deviation of the distribution is  

SD = σ = (b – a) / sqrt(12)

= (46 – 25) / sqrt(12)

= 21/3.464102

= 6.062minutes

SD = σ = 6.062minutes

Now, we have to find P(30<X<40)

P(30<X<40)

= (40 – 30) / (46 – 25)

= 10/19

= 0.4762

Required probability = 0.4762

Now, we have to find the value of a for which P(X≤a) = 0.9

We have, P(X≤a)

= (a – 25) / (46 – 25)

= (a – 25) / 21

(a – 25) / 21 = 0.9

(a – 25) = 0.9*21

So, a = 18.9 + 25 = 43.

a = 43.9

= 43.9minutes

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

the answer is 2/54

step-by-step explanation: