Answer:a)
b) theta = 45°
c) S = 84.375m
Step-by-step explanation: see attachment below
Answer:
0.1994 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 166 pounds
Standard Deviation, σ = 5.3 pounds
Sample size, n = 20
We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =

P(sample of 20 boxers is more than 167 pounds)
Calculation the value from standard normal z table, we have,
0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds
80 inches or feet.. I think you are suppose to multiply 40x2 sorry if im wrong
Answer:
The probability that a randomly selected programmer major received a salary less than 38000 is 0,3085
Step-by-step explanation:
We will assume that the salaries are Normally distributed. Lets call X the salary of a random major programmer in dollars. We want the pprobability of X being less than 38000. For it, we will standarize X. Lets call W the standarization, given by the formula

Lets denote
the cumulative distribution function of the standard normal variable W. The values of
are well known and they can be found in the attached file. Now, lets calcualte the probability of X being less than 38000 using

Since the density function of a standard normal random variable is symmetric, then 
The probability that a randomly selected programmer major received a salary less than 38000 is 0,3085.
The Answer To Your Question Would Be 12