Ask question

Ask question

All categories

Vera_Pavlovna [14]

3 years ago

11

suppose you are selected for a survey, but your friends persuade you to respond in a way that you dont actyally agree with. what

is this an expample of?

1 answer:

sdas [7]3 years ago

3 0

Answer:

peer pressure

Step-by-step explanation:

You might be interested in

HELP ME PLEASEEEEEEEEEEE

AlexFokin [52]

Answer: T. F. F. T.

Step-by-step explanation:

IGJ

8 0

3 years ago

Read 2 more answers

What is the area of this partial circle with a radius of 14 inch

AveGali [126]

Step-by-step explanation:

Since this circle is partial we will divide the area of the whole circle by 2

the area of a circle is expressed by $\pi r^2$ so insert 14 for the radius.

$14^2\pi$

$196\pi$

divide the area by 2

$98\pi=307.876$

Hope that helps :)

6 0

3 years ago

9x-22=23+4x I don't understand how to do this

anygoal [31]

9x-22=23+4x Now you subtract 4x from both sides to get the variable on one side.. 5x-22=23 Now solve as you would solve a 2 step equation. 5x=45 Divide both sides by 5 to get the variable by itself. x=9

4 0

2 years ago

ratelena [41]

The total number of calories for 18 cups of milk is 1620 calories.
If two cups of milk is 180 calories then divide 180/2 to find the unit rate.
Then multiply 90 by the number of cups (18) and then you get the total number of calories for that many drinks.

6 0

2 years ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

$Z = \frac{X - \mu}{s}$

X = 205

$Z = \frac{X - \mu}{s}$

$Z = \frac{205 - 200}{5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{195 - 200}{5}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

$Z = \frac{X - \mu}{s}$

$Z = \frac{210 - 200}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{190 - 200}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0

3 years ago