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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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What ratio is equvalent to 6 over 12​
Ber [7]

Hello from MrBillDoesMath!

Answer:

1/2

Discussion:

6/12 =                            => as 12 = 6*2

6 /( 6*2)  =                      => cancel 6's in num and denom.

1/2

Thank you,

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A sixth grade class of 295 students is having an end of the year laser tag party. The party will cost
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3 years ago
The mean number of daily surgeries at a local hospital is 6.2. Assume that surgeries are random, independent events. (a) The cou
Scilla [17]

Answer:

a) correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

b) probability for only 2 or less than 2  surgeries  in a given day is 0.0536

Step-by-step explanation:

Given data:

mean number is given as 6.2

correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

we know that variance is given as\sigma^2 = mean = 6.2

hence, standard deviation is given as \sqrt{6.2} = 2.48998

thus standard deviation is 2.49

correct option is

Poisson distribution is 6.2 and standard deviation is 2.49

b) probability for only 2 or less than 2  surgeries  in a given day is

P(X \leq 2) = P(X =0) + P(X =1) + P(X =2)

                 = \frac{e^{-6.2} 6.2^0}{0!} +\frac{e^{-6.2} 6.2^1}{1!} +\frac{e^{-6.2} 6.2^2}{2!}

                 = 0.002029 + 0.012582+ 0.039006

                 = 0.053617

                = 0.0536

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