There are three random variables: X ~ Normal with mean = 60 and Standard deviation = 35 Z ~ Normal Standard W ~ t(18)df Calculate in R, the following: 1) P(X>66) = Answer 2) Find "A", such that: P(X Then A = Answer 3) P(Z>-1.54) = Response 4) Find "B", such that: P(W>B) = 0.25 ==> Then B= Answer USE R STUDIOS. DO NOT ROUND DECIMALS.

The random variable x has a normal distribution with mean 50 and variance 9. Find the value of x, call it x0, such that: a) P(x ≤ xo) = 0.8413 b) P(x > xo) = 0.025 c) P(x > xo) = 0.95 d) P(41 ≤ x ≤ xo) = 0.8630

Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.25 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 36. Complete part (a) through (c) below. ..... a. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 36. The value of the mean is u = рeas. (Type an integer or a decimal. Do not round.)