normal distribution (motorists' mean driving speed, etc)

1) Given: X ~ N(A, B^2), P(X < 4.1) = 0.75, and P(X < 5.2) = 0.95, find P(2.5 < X < 3)

I got (4.1 - A)/B = 0.7 and (5.2 - A)/B = 1.675, but dont know where to go, or what I'm supposed to do.

2) Ninety percent of motorists drive at speeds greater than 100km/h; only 5 percent drive at less than 95 km/h. If speeds are assumed to be normally distributed, determine the motorists' mean driving speed and standard deviation.

I have the same work here for the above problem, but it leads nowhere.

You're on the right track.

mcrae said:

1) Given: X ~ N(A, B^2), P(X < 4.1) = 0.75, and P(X < 5.2) = 0.95, find P(2.5 < X < 3)

I got (4.1 - A)/B = 0.7 and (5.2 - A)/B = 1.675, but dont know where to go, or what I'm supposed to do.

Click to expand...

Solve your two equations for A and B. Then find P( (2.5-A)/B < (X-A)/B < (3-A)/B ).

2) Ninety percent of motorists drive at speeds greater than 100km/h; only 5 percent drive at less than 95 km/h. If speeds are assumed to be normally distributed, determine the motorists' mean driving speed and standard deviation.

I have the same work here for the above problem, but it leads nowhere.

Click to expand...

Similar but just solve for A and B.

Have you tried solving for A and B? You have two equations. It should work. You haven't forgotten your algebra, have you? :wink:

I have. So did the rest of my class. We stink at this.

Thank you!

Ouch! OK, then one for free.

The Substitution Method
(4.1 - A)/B = 0.7 and (5.2 - A)/B = 1.675

B ≠ 0

(4.1 - A) = 0.7*B and (5.2 - A) = 1.675*B
4.1 - 0.7*B = A and 5.2 - A = 1.675*B
4.1 - 0.7*B = A and 5.2 - (4.1 - 0.7*B) = 1.675*B
5.2 - 4.1 + 0.7*B = 1.675*B
1.1 = 1.675*B - 0.7*B
1.1 = 0.975*B
B = 1.1/0.975 = 1.128205
A = 4.1 - 0.7*B = 4.1 - 0.7*(1.128205) = 3.310256

Checking.
(4.1 - A)/B = 0.7??
(4.1 - 3.310256)/1.128205 = 0.700000443 -- Check

(5.2 - A)/B = 1.675??
(5.2 - 3.310256)/1.128205 = 1.675000554 -- Check

Done.