Show that p(y) p(y − 1) = (n − y + 1) p yq > 1 if y < (n + 1)p. This establishes that p(y) > p(y − 1) if y is small (y < (n + 1)p) and p(y) < p(y − 1) if y is large (y > (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on. (n + 1)p > y

Question is not well presented

Consider the binomial distribution with n trials and P(S) = p.

Show that

p(y) /p(y − 1) = (n − y + 1)p /yq > 1 if y < (n + 1)p.

This establishes that p(y) > p(y − 1) if y is small (y < (n + 1)p) and p(y) < p(y − 1) if y is large (y > (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on.

See Explanation Below

Step-by-step explanation:

Given that

p(y) /p(y − 1) = (n − y + 1)p /yq > 1 if y < (n + 1)p.

First, we make the following assumption

p(y) /p(y − 1) = (n − y + 1)p /yq = 1;

So, we have

(n − y + 1)p /yq = 1

(n − y + 1)p = yq

Note that p + q = 1;.

So, q = 1 - p

Substitute 1-p for q in the above expression

(n − y + 1)p = y(1-p)

np - py + p = y - py

Solve for y

Collect like terms

np + p = y - py + py

np + p = y;

So, y = np + p

y = (n + 1)p

From the above,

We know that

p(y) /p(y − 1) = 1

if y = (n + 1)p.

Similarly, we've also obtain that

p(y) /p(y − 1) > 1 if y < (n + 1)p.

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(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240