Do questions 1 and 2 (View picture for question 1 )

Finish the pattern 18, 21, _ , 24, _ , 30

omeli [17]

Goes +3
18, 21, 24, 27, 30

4 0

3 years ago

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

(a) The distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b) The value of $P(\bar X$ is 0.7642.

(c) The value of $P(\bar X\geq 69.1)$ is 0.3670.

Step-by-step explanation:

A random sample of size n = 10 is selected from a population.

Let the population be made up of the random variable X.

The mean and standard deviation of X are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. n = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$

And the standard deviation of the distribution of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74$

Thus, the distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b)

Compute the value of $P(\bar X$ as follows:

$P(\bar X$

$=P(Z$

*Use a z-table for the probability.

Thus, the value of $P(\bar X$ is 0.7642.

(c)

Compute the value of $P(\bar X\geq 69.1)$ as follows:

Apply continuity correction as follows:

$P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)$

$=P(\bar X>69.6)$

$=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})$

$=P(Z>0.34)\\=1-P(Z$

Thus, the value of $P(\bar X\geq 69.1)$ is 0.3670.

7 0

3 years ago

Suppose in the population, the Anger-Out score for men is two points higher than it is for women. The population variances for m

tatuchka [14]

Answer:

a)= 2

b) 6.324

c) P= 0.1217

Step-by-step explanation:

a) The mean of the sampling distribution of X`1- X`2 denoted by ux`-x` = u1-u2 is equal to the difference between population means i.e = 2 ( given in the question)

b) The standard deviation of the sampling distribution of X`1- X`2 ( standard error of X`1- X`2) denoted by σ_X`1- X`2 is given by

σ_X`1- X`2 = √σ²/n1 +σ²/n2

Var ( X`1- X`2) = Var X`1 + Var X`2 = σ²/n1 +σ²/n2

so

σ_X`1- X`2 =√20 +20 = 6.324

if the populations are normal the sampling distribution X`1- X`2 , regardless of sample sizes , will be normal with mean u1-u2 and variance σ²/n1 +σ²/n2.

Where as Z is normally distributed with mean zero and unit variance.

If we take X`1- X`2= 0 and u1-u2= 2 and standard deviation of the sampling distribution = 6.324 then

Z= 0-2/ 6.342= -0.31625

P(-0.31625<z<0)= 0.1217

The probability would be 0.1217

8 0

2 years ago

Which four points are coplanar

jenyasd209 [6]

Answer:

B, A, F, H

Step-by-step explanation:

8 0

3 years ago

Lock 1: Write your response in Alphabetical Order and in ALL CAPITAL LETTERS (Do not put any spaces or commas.) You must choose

Dennis_Churaev [7]

Answer:

I believe A C and E.

7 0

3 years ago