Question

The spot speeds at a particular location are normally distributed with a mean of 51.7kmph and a standard deviation of 8.3 kmph what is the probability that a)the speeds exceed 65 kmph,the speeds lie between 40 kmph and 70kmph,what is the 85th percentile speed

Answer 1

(i) the speeds exceed 65 kmph (2.5)

(ii) Speeds lie between 40 kmph and 70 kmph (2.5)

(iii) 85th percentile speed. (3)

(i) Speeds exceed 65 kmph

P(x>65) = 1-P(x<65)

Standard variate = 65−51.7

8.3

= 1.6

P(x>65) = 1-P(x<65) = 1-0.952 = 4.8 %

(ii) Speeds lie between 40 kmph and 70 kmph

40 =

40 − 51.7

8.3

= −1.41

70 =

70 − 51.7

8.3

= 2.21

P (1- ɸ (1.41) = 1- 0.9207 =0.0793

P(40 < x < 51.7) = 0.5-0.0793 = 0.4207

P(51.7 < x < 70) = 0.9864 - 0.5 = 0.4864

P(40 < x < 70) = 0.4207+ 0.4864 = 90.71%

(iii) 85th percentile speed

ɸ (Z) = 0.85 for which Z=1.04

Answer 2

85 percentile speed is 1.04 kmph.

Explanation:

Given:

To find:

What is the probability that

a) the speeds exceed 65kmph

b) the speeds lie between 40kmph and 70kmph

c) what is the 85th percentile speed

Solution:

a) probability that the speeds exceed 65kmph

Standard Variate for 65kmph $=\frac{65-\;51.7}{8.3} = 1.6$

$P(x>65)\;=\;1-P(x<65) = 1-0.952 = 4.8\%$

probability that the speeds exceed 65kmph is $4.8\%$

b) probability that the speeds lie between 40kmph and 70kmph

Standard Variates are $\frac{40-\;51.7}{8.3} = -1.41$ (for 40kmph) and $=\frac{70-\;51.7}{8.3} = 2.21$ (for 70kmph)

$P\;(1-\;\phi\;(1.41))\;=\;1-\;0.9207\;=0.0793$

$P(40\;<\;x\;<\;51.7)\;=\;0.5-0.0793\;=\;0.4207$

$P(51.7\;<\;x\;<\;70)\;=\;0.9864\;-\;0.5\;=\;0.4864$

$P(40\;<\;x\;<\;70)\;=\;0.4207+\;0.4864\;=\;0.9071=\;90.71\%$

probability that the speeds lie between 40kmph and 70kmph is $90.71\%$

c) what is the 85th percentile speed

$\phi\;(Z)\;=\;0.85\;$ for which $Z=1.04$

the 85th percentile speed is 1.04kmph.