All categories

2 years ago

The z score boundaries for the critical region are determined by

1 answer:

4 0

<span> The boundaries for the critical region determine how much distance between the sample mean and population mean is needed to reject the null. </span>

You might be interested in

Loki's home repairs will cost him $1,200. The bank will give him a loan for the full amount at an interest rate of 21%. The bank

bogdanovich [222]

Formula of interest =PRT divided by 100

6 0

2 years ago

Convert 1/4 inch into millimeters

Likurg_2 [28]

1 in equals 25.4mm exactly so

.25in(25.4mm/in)=6.35mm

6 0

2 years ago

Read 2 more answers

Mindy's dog, Rex, weighed 80 pounds a year ago. Now he weighs 92 pounds. What is the percent increase in his weight?

nadya68 [22]

Answer:

15%

Step-by-step explanation:

To find the percent increase, use the equation:

percent of increase=<u>amount of increase</u>

original amount

The amount of increase is 92-80 = 12. The original amount of 80; this gives us

12/80 = 0.15 = 15%

5 0

2 years ago

See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly se

REY [17]

Answer:

a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b) 0.0064% probability that all of the selected adults believe in reincarnation.

c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d) Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

Step-by-step explanation:

For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 6, p = 0.2$

a. What is the probability that exactly five of the selected adults believe in reincarnation?

This is P(X = 5).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015$

There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b. What is the probability that all of the selected adults believe in reincarnation?

This is P(X = 6).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064$

There is a 0.0064% probability that all of the selected adults believe in reincarnation.

c. What is the probability that at least five of the selected adults believe in reincarnation?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564$

There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?

5 is significantly high if $P(X \geq 5) < 0.05$

We have that

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05$

Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

5 0

2 years ago

A sandwich shop offers ham, turkey, tuna, chicken salad, and roast beef. It has Swiss, American, and provolone cheese. You can o

a_sh-v [17]

5 different meats

3 different cheeses

3 different breads

5x3x3 = 15*3 = 45

there are 45 choices

4 0

2 years ago