Answer to Question #120153 in Statistics and Probability for que

Question

The amount of time that a vehicle spends in a petrol bunk is a random variable with mean, µ
= 4.5 minutes and a standard deviation, σ = 1.8 minutes. Find the probability that on an average
day, the time spend at the petrol bunk is
(a) at most 3.6 minutes
(b) more than 5.5 minutes
(c) at least 3.2 minutes but less than 4 minutes
Compute the value of time spend at the petrol bunk for
(d) 20% of the lower end
(e) 10% of the higher end

Answer 1

Let X = the random variable denoting the amount of time that a vehicle spends in a petrol bunk

When the distribution is not mentioned, it is conventionally assumed to be Normal.

Here the distribution of X has not been mentioned, so we assume

X ~ N(µ = 4.5, σ² = 1.8²)

Then we have,

Z = $\frac{X-\mu}{\sigma}$ ~ N(0, 1), Z is the standard normal variate

(a) The probability that on an average day, the time spend at the petrol bunk is at most 3.6 minutes

= P(X $\leq$ 3.6)

= P($\frac{X-4.5}{1.8}\leq\frac{3.6-4.5}{1.8}$)

= P(Z $\leq$ - 0.5)

= $\Phi$(- 0.5)

= 0.3085

Answer: The probability that on an average day, the time spend at the petrol bunk is at most 3.6 minutes is 0.3085.

(b) The probability that on an average day, the time spend at the petrol bunk is more than 5.5 minutes

= P(X > 5.5)

= P($\frac{X-4.5}{1.8}>\frac{5.5-4.5}{1.8}$)

= P(Z > 0.56)

= 1 - P(Z $\leq$ 0.56)

= 1 - $\Phi$(0.56)

= 1 - 0.7123

= 0.2877

Answer: The probability that on an average day, the time spend at the petrol bunk is more than 5.5 minutes is 0.2877.

(c) The probability that on an average day, the time spend at the petrol bunk is at least 3.2 minutes but less than 4 minutes

= P(3.2 $\leq$ X $<$ 4)

= P($\frac{3.2-4.5}{1.8}​\leq\frac{X-4.5}{1.8}<\frac{4-4.5}{1.8}​$)

= P(- 0.72 $\leq$ Z $<$ - 0.28)

= $\Phi$(- 0.28) - $\Phi$(- 0.72)

= 0.3897 - 0.2358

= 0.1539

Answer: The probability that on an average day, the time spend at the petrol bunk is at least 3.2 minutes but less than 4 minutes is 0.1539.

(d) Let t₁ be the value of time spend at the petrol bunk for 20% of the lower end

Then P(X < t₁) = 20%

i.e. P($\frac{X-4.5}{1.8}<\frac{t_1-4.5}{1.8}$) = 0.2

i.e. P(Z < $\frac{t_1-4.5}{1.8}$) = 0.2

i.e. $\Phi$($\frac{t_1-4.5}{1.8}$) = 0.2 = $\Phi$(-0.84)

i.e. $\frac{t_1-4.5}{1.8}$ = -0.84

i.e. t₁ = -2.988 $\approx$ 3

Answer: The value of time spend at the petrol bunk for 20% of the lower end is 3 minutes.

(e) Let t₂ be the value of time spend at the petrol bunk for 10% of the higher end

Then P(X > t₂) = 10%

i.e. P($\frac{X-4.5}{1.8}>\frac{t_2-4.5}{1.8}$ ​) = 0.1

i.e. 1 - P($\frac{X-4.5}{1.8}\leq\frac{t_2-4.5}{1.8}$) = 0.1

i.e. P(Z $\leq\frac{t_2-4.5}{1.8}$) = 1 - 0.1 = 0.9

i.e. $\Phi(\frac{t_2-4.5}{1.8}$) = 0.9 = $\Phi$(1.28)

i.e. $\frac{t_2-4.5}{1.8}$ ​ = 1.28

i.e. t₂ = 6.804 $\approx$ 6.8

Answer: The value of time spend at the petrol bunk for 10% of the higher end is 6.8 minutes.