Percentages within intervals ( ..., ...)

[shamieh]

a study carried out to investigate the distribution of total braking time (reaction time plus accelerator to brake movement time in ms) during real driving conditions at 60 km/hr gave the following summary information on the distirbution of times ("A Field Study on Braking Responses During Driving" Ergonomics, 1995: 1903-1910: mean = 535, s = 80. Assume the braking times have approximately a bell shaped distribution.

a) approximately what percentage of braking times are in the interval (535,615)
b)what value do the top 2.5% of braking times exceed?

I know this is relatively simple I just don't know the formulas or steps to get the solutions.

 

[MarkFL]

You are told yu have normally distributed data where:

\(\displaystyle \mu=535,\,\sigma=80\)

a) In order to use your table, you will need to standardize the given data values, using:

\(\displaystyle z=\frac{x-\mu}{\sigma}\)

What do you get when you convert the given raw data of 535 and 615 into $z$-scores?

Do you understand what to do with the table once you have the $z$-scores? It can differ depending on whether the two $z$-scores have the same sign or not.

 

[shamieh]

ok so: $z = \frac{615 - 535}{80}$ correct? -> 1 standard deviation from the mean which is 34.1%

 

[MarkFL]

ok so: $z = \frac{x - 535}{80}$ correct?

Yes, and using that you need to standardize the data $x=535,\,615$. Standardizing gives us the number of standard deviations a datum is from the mean. The sign is a consequence of the datum being either less than or greater than the mean.

 

[shamieh]

No, I'm not sure how do I use the table for part B

 

[MarkFL]

Just so we're clear on part a), we have:

\(\displaystyle z_1=\frac{535-535}{80}=0\)

\(\displaystyle z_1=\frac{615-535}{80}=1\)

Since a table will typically give the area under the curve between the mean and some positive $z$-score, we simply read from the table to get:

\(\displaystyle P(x)\approx34.13\%\)

Now, for part b), we want to first find the $z$-score with an area of 0.975 to its left. We know an area of 0.5 is to the left of the mean, so we want the $z$-score associated with an area of 0.475...then we want to convert this $z$-score into raw data.

Can you proceed?

 

[shamieh]

so I did 50%-2.5% to get 47.5% and then used the table to get $1.906$ for my $z$

then plugged them in and solved for $x$ to get 687.48

$1.906$ = $\frac{x-535}{80}$

$x = 687.48$

 

[MarkFL]

You made a slight error with your table...the $z$-score you want is 1.96...so you have:

\(\displaystyle x=691.8\)