Suppose that x has a binomial distribution with n = 201 and p = 0.45. (Round np and n(1-p) answers to 2 decimal places. Round your answers to 4 decimal places. Round z values to 2 decimal places. Round the intermediate value (σ) to 4 decimal places.) (a) Show that the normal approximation to the binomial can appropriately be used to calculate probabilities about x.

a) It can be used because np and n(1-p) are both greater than 5.

Step-by-step explanation:

Binomial distribution and approximation to the normal:

The binomial distribution has two parameters:

n, which is the number of trials.

p, which is the probability of a success on a single trial.

If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.

In this question:

So, lets verify the conditions:

np = 201*0.45 = 90.45 > 5

n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5

Since both np and n(1-p) are greater than 5, the approximation can be used.

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation: