Here's an example of solving for dy/dx using implicit differentiation


y^2 + 2xy^3 - 3x^2 = 14

2y(dy/dx) + 2x(3y^2)(dy/dx) + 2y^3 - 6x = 0

2y(dy/dx) + 6xy^2(dy/dx) = 6x - 2y^3

dy/dx(2y +6xy^2) = 6x - 2y^3

dy/dx = (6x-2y^3)/(6xy^2 + 2y)

dy/dx = (3x-y^3)/(3xy^2 + y)

For Normal distribution with mean μ = 400
and standard deviation σ = 100 find probability
that x is between 340 and 714: P( 340 < x < 714)

0.3427

0.5462

0.7249

0.9140

We have mean = 400 and standard deviation of 100, now we want P(X > 619), Z = (619 - 400)/100 = 2.19

Z(2.19) = .9857, so we take 1- .9857 since the probability is greater than, so we get .0143
For the next question, mean is 400 and standard deviation is 100, want: P( 340 < x < 714)
Z = (340 - 400)/100 = -0.6 and Z = (714 - 400)/100 = 3.14

So now we find Z(3.14) and Z(-0.6) and subtract them to get the probability in between. .9992 - .2743 = .7249

A quick note:

For the test just comparing two means when population standard deviation is not known and sample size is small, you can do a t-test . The formula for the test statistic is t = (x1-bar - x2-bar)/sqrt(s1^2/n1 +s2^2/n2), then compare to the critical value t , found using any t-distribution chart.

we calculate Chi-square by getting the sum(Observed - Expected)^2/(Expected)



You get the expected in each cell by taking row total times column total divided by overall total
Expected Low Drunk Driving and Low Under Age is (22)(20)/50 = 8.8
Expected High Drunk Driving and Low Under Age is (28)(20)/50 = 11.2
Expected Low Drunk Driving and High Under Age is (22)(30)/50 = 13.2
Expected High Drunk Driving and High Under Age is (28)(30)/50 = 16.8

so Chi-square statistic is (8-8.8)^2/8.8 + (14-13.2)^2/13.2 + (12 -11.2)^2/11.2 + (16- 16.8)^2/16.8 = 0.216

Now we get the p-value, with df of (row-1)(column - 1). There are 2 rows and 2 columns in the contingency table, so df = 1. I use this site.... http://www.socscistatistics.com/pvalues/chidistribution.aspx
The p-value is 0.642105. Since the p-value > .05