Question

The reaction time of a driver to visual stimulus is normally distributed with a mean of 0.4 seconds and a standard deviation of 0.05 seconds.(a) What is the probability that a reaction requires more than 0.5 second? (b) What is the probability that a reaction requires between 0.4 and 0.5 second? (c) What is the reaction time that is exceeded 90% of the time?

Answer 1

a) 0.0228

b) 0.4772

c) 0.336

Explanation:

Mean(μ) = 0.4 seconds

Standard deviation (σ) = 0.05 seconds

From normal distribution,

Z= (x - μ) / σ

a) P(x > 0.5)

Let x be the random variable for the required seconds

When x= 0.5

Z = (0.5 - 0.4)/0.05

Z = 2

From the normal distribution table, 2= 0.4772

φ(Z) = 0.4772

Recall that when Z is positive,

P(x >a) = 0.5 - φ(Z)

P(x > 0.5) = 0.5 - 0.4772

= 0.0228

b) For x = 0.4

Z= (x - μ) / σ

= (0.4 - 0.4) / 0.05

= 0

For x= 0.5

Z= (x - μ) / σ

= (0.5 - 0.4) / 0.05

= 2

From the table, P(0.4 < x < 0.5) = P(0 < Z < 2)

So we have

P(Z < 2) - P(Z<0)

From the table, 2 = 0.4772 and 0 = 0

We then have

0.4772 - 0

= 0.4772

c) we are looking for x such that 90% of the values lie above it or 10% of the value lie below it.

From the table , 10% probability gives a z value of -1.28

x = μ + Zσ

x = 0.4 + (-1.28*0.05)

x = 0.4 - (1.28*0.05)

= 0.336

Answer 2

a)
`P(X>0.5)=P((X-\mu)/(\sigma)>(0.5-\mu)/(\sigma))=P(Z>(0.5-0.4)/(0.05))=P(z>2)`

`P(z>2)=1-P(z<2)`

`P(Z>2) = 1-P(Z<2)= 1- 0.97725=0.02275`

b)
`P(0.4<X<0.5)=P((0.4-\mu)/(\sigma)<(X-\mu)/(\sigma)<(0.5-\mu)/(\sigma))=P((0.4-0.4)/(0.05)<Z<(0.5-0.4)/(0.05))=P(0<z<2)`

`P(0<z<2)=P(z<2)-P(z<0)`

`P(0<z<2)=P(z<2)-P(z<0)=0.97725-0.5=0.47725`

c)
`a=0.4 +1.28*0.05=0.464`

So the value of height that separates the bottom 90% of data from the top 10% is 0.464.

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the reaction time of a driver to visual stimulus of a population, and for this case we know the distribution for X is given by:

`X \sim N(0.4,0.05)`

Where
`\mu=0.4` and
`\sigma=0.05`

We are interested on this probability

`P(X>0.5)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X>0.5)=P((X-\mu)/(\sigma)>(0.5-\mu)/(\sigma))=P(Z>(0.5-0.4)/(0.05))=P(z>2)`

And we can find this probability using the complement rule:

`P(z>2)=1-P(z<2)`

And using the normal standard table or excel we have this:

`P(Z>2) = 1-P(Z<2)= 1- 0.97725=0.02275`

Part b

`P(0.4<X<0.5)=P((0.4-\mu)/(\sigma)<(X-\mu)/(\sigma)<(0.5-\mu)/(\sigma))=P((0.4-0.4)/(0.05)<Z<(0.5-0.4)/(0.05))=P(0<z<2)`

And we can find this probability on this way:

`P(0<z<2)=P(z<2)-P(z<0)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(0<z<2)=P(z<2)-P(z<0)=0.97725-0.5=0.47725`

Part c

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.10` (a)

`P(X<a)=0.90` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.10 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.9`

`P(z<(a-\mu)/(\sigma))=0.9`

But we know which value of z satisfy the previous equation so then we can do this:

`z=1.28=(a-0.4)/(0.05)`

And if we solve for a we got

`a=0.4 +1.28*0.05=0.464`

So the value of height that separates the bottom 90% of data from the top 10% is 0.464.