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Ne4ueva [31]
3 years ago
10

A simple random sample of size nequals57 is obtained from a population with muequals69 and sigmaequals2. Does the population nee

d to be normally distributed for the sampling distribution of x overbar to be approximately normally​ distributed? Why? What is the sampling distribution of x overbar​?
Mathematics
1 answer:
dusya [7]3 years ago
6 0

Answer:

The population does not need to be normally distributed for the sampling distribution of \bar{X} to be approximately normally distributed. Because of the central limit theorem. The sampling distribution of \bar{X} is approximately normal.

Step-by-step explanation:

We have a random sample of size n = 57 from a population with \mu = 69 and \sigma = 2. Because n is large enough (i.e., n > 30) and \mu and \sigma are both finite, we can apply the central limit theorem that tell us that the sampling distribution of \bar{X} is approximatelly normally distributed, this independently of the distribution of the random sample. \bar{X} is asymptotically normally distributed is another way to state this.

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In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there
yan [13]

Answer:

Explained below.

Step-by-step explanation:

According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normally.  

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

\mu_{\bar x}=\mu

And the standard deviation of the sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

a

The expected value of the sample mean of their weights is same as the population mean, <em>μ</em> = 1515 lbs.

b

The standard deviation of the sampling distribution of the sample mean weight is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{25}{\sqrt{16}}=6.25

c.

The average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2500 lbs. is:

\text{Average Weight}=\frac{2500}{16}=156.25

d

Compute the probability that a random sample of 16 persons on the elevator will exceed the weight limit as follows:

P(\bar X > 156.25)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{156.25-151}{6.25})\\\\=P(Z>0.84)\\\\=1-P(Z

4 0
3 years ago
C(n)=-6+5(n-1) what is the 8th term in the sequence
Natasha_Volkova [10]

Answer:

  29

Step-by-step explanation:

Put 8 where n is, then do the arithmetic.

  C(8) = -6 +5(8 -1) = -6 +5(7) = -6 +35

  C(8) = 29

5 0
3 years ago
he original price of a tractor was $45,000. The price of the tractor decreases at a steady rate of 8.3% per year. What is the va
Debora [2.8K]
  • Answer:

<em>$29178.25</em>

  • Step-by-step explanation:

<em>x = p(1 - r)ⁿ</em>

<em>p = original price</em>

<em>r = 8.3% = 0.083</em>

<em>n = 5 years</em>

<em>x = $45000(1 - 0.083)⁵</em>

<em>= $45000ₓ0.917⁵</em>

<em>= $45000ₓ0.64840548256</em>

<em>= </em><em>$29178.25</em>

5 0
4 years ago
What is the answer to this algebra?<br>5ab + 3cb - 2ba - 2bc<br>​
Nookie1986 [14]

Answer:b x (3a+c)

Step-by-step explanation:

Collect like terms

3ab + 3cb-2bc

3ab+cb

Factor out b from the expression

b x(3a+c)

7 0
3 years ago
At First Class Pizza, 12% of the pizzas made yesterday were pepperoni pizzas. If 27 pizzas have pepperoni, how many pizzas were
gogolik [260]

Answer: 225

Step-by-step explanation:

Let the total number of pizza made be represented by x.

Therefore, the situation in the question can be written as:

12% × x = 27

12/100 × x = 27

0.12 × x = 27

0.12x = 27

Divide both side by 0.12

0.12x/0.12 = 27/0.12

x = 225

The total pizza made was 225.

3 0
3 years ago
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