Question

2. A radar unit is used to measure the speed of automobile on an expressway during rush-hour traffic. The speeds of individual automobiles are normally distributed with a standard deviation of 4 mph, a. Find the mean of all speeds if 3% of the automobiles travel faster than 72 mph. Round the mean to the newest forth b. Using the mean you found in part a, find the probability that a car is traveling between 70 mph and 75 mph. Interpret the meaning of this answer c. Using the mean you found in part a, find the 25th percentile for the variable speed".

Answer 1

şolution :- la Given that Standard deviation - 4 mph. (@ tere we have to calculate the mean of all 34. Of the automobiles travel faster than 72 mph. .: The box population coas normally distributed 34. MOCK 1 272 fuu. Mean speed NOOD Finding the corresponding E-value in Z-table for (1-0.3 = 0.97) → 977 on the right side of z-table , We get 20.9699 = 1.88 Hence The equation for population Z-valve is

quo , We got Substituting the above values in ..72-01 → 1.88 =- 4 X 1.88 = 42- ei=64.48 Tee - 64.5| (b) Here 64.5 10 175 NOED Z- value at yo mph 270 = 70-'64-5 = 1:38 = 0,9162 Z-Value at 75 mph 7 7.75 = 75-64.5 = 2.63 = 0.9957

tence The probability of vehicles travelling betegeen to and 15 mph cao 7 45 - 740 = 0.9957-0.9162 = 0.0795 1775 - 740 = 0.08 :: 7.95 %. of vehicles are travelling at a Speed of Yo and 75 mph C) NO CO The 15th peocentile for the variable 'speed " is olul (25th percentile) en the Z- table , We found out the 251. respective probability and the values comes out to be - 0.68 Substituting the above value in equco, we get

X-64.5 - = -0.68 X - 645 - 4X(-0.68) x = 61.76 Tx~ 61.8| *** Thanking you --***

Answer 2

SOLUTION :

a.

Let mean be m mph.

So,

P(x > 72) = 3% = 0.03

P(z > (72 - m)/4 = 0.03

From ND table, for P(x > 72) = 0.03,   z = 1.8814

So,

(72 - m)/4 = 1.8814

=> m = 72 - 4*1.8814 = 64.4744 = 64.47  mph (ANSWER).

b.

P(x between 70 and 75)

= P(z between (70 - 64.4744)/4  and (75 - 64.4744)/4)

= P(z between 1.3814 and 2.6314)

= P(z ≤ 2.6314) - P(z ≤ 1.3814P

= 0.99577 - 0.9164

= 0.07937 = 0.0794(ANSWER)

This means 7.94% automobiles travel between 70 and 75 mph. (ANSWER)

c.

25th percentile means top 25% are below the cutoff value of ‘a’ mph).

So,

P(x < a) = P(z < (a - 64.4744)/4) = 0.25

From ND table the cutoff z = - 0.6791

=> (a - 64.4744)/4 = - 0.6791

=> a = - 0.6791*4 + 64.4744

=> a = 61.758 = 61.76 mph 

So, 25th percentile is 61.76 mph (ANSWER).