Question

• Suppose reaction time for drag racers is know to be on average 50 ms with a standard deviation of 10 ms and that reaction time is normally distributed.

1. • What is the reaction time that separates the fastest 10% of racers from the rest?
2. What is the interval that contains the middle 95% of racers?
3. Above what value would you expect to find the slowest 60%?

Can you please explain in detail? plus i know for this problem we have to use this formula X= z x standard deviation + mean. Where do i get the z for this problem? i know the mean and Standard deviation but where am i supposed to get the Z?

Answer 1

Solution:

Given: Reaction time for drag racers is know to be on average 50 ms with a standard deviation of 10 ms and that reaction time is normally distributed.

That is:  

Part a) What is the reaction time that separates the fastest 10% of racers from the rest?

Fastest reaction time means it should take minimum time than others.

That is we have to find x value such that:

P( X < x ) = 10%

P( X < x ) = 0.10

To find this x value , we need to find z value such that:

P( Z < z ) = 0.10

Look in z table for area = 0.1000 or its closest area and find z value:

Area 0.1003 is closest to 0.1000 and it corresponds to -1.2 and 0.08

thus z = -1.28

Now use following formula to find x value:

Thus  the reaction time that separates the fastest 10% of racers from the rest is 37.2 ms.

Part b) What is the interval that contains the middle 95% of racers?

That is find two x values such that:

P( x1 < X < x2) =95%

Thus to find these x1 and x2 values, find z values such that:

P( -z < Z < z )= 95%

P( -z < Z < z )= 0.95

P( Z < z) - P( Z < -z) = 0.95

Since middle area is 0.95, then rest area in two tails = 1 - 0.95 = 0.05

We divide this 0.05 area in two tails equally , that is : 0.05/2 = 0.025 area in left tail and 0.025 area in right tail

that is: P( Z < -z ) = 0.025 and P( Z > z ) = 0.025

Thus to find -z value, look in z table for Area = 0.025 and find corresponding z value.

Area 0.0250 corresponding to -1.9 and 0.06

that is : - z = -1.96

then z = 1.96

Thus we get:

P( Z < -1.96 ) = 0.025 and P( Z > 1.96) = 0.025

Now use z = 1.96 to find x₁ and x₂ values:

( For lower x value we use -z value and for upper x value we use +z value)

and

Thus the interval that contains the middle 95% of racers are between and .

Part c) Above what value would you expect to find the slowest 60%?

That is we have to find x value such that:

P( X > x ) = 60%

Thus find z such that:

P( Z > z) = 60%
P( Z > z) = 0.60

That is find :

P(Z < z ) =1 - P( Z > z)

P(Z < z ) =1 - 0.60

P(Z < z ) = 0.40

Thus look in z table for Area = 0.4000 or its closest area and find z value.

Area = 0.4013 is closest to 0.4000 and it corresponds to -0.2 and 0.05

Thus z = -0.25

Now use following formula to find x value:

Thus above 47.5 ms , we would expect to find the slowest 60%.