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MAVERICK [17]
3 years ago
8

Suppose a large population has a mean μand standard deviation σ, and a simple randomsample of size nis taken. The sampling distr

ibution of the sample mean xhas mean andvariance respectively equal to(a)μ/nand σ2/n.(b)μand σ/n.(c)μ/nand σ2/n2.(d)μand σ2/n.
Mathematics
1 answer:
My name is Ann [436]3 years ago
4 0

Answer:

(d) μ and σ²/n

Step-by-step explanation:

In a sampling distribution of sample means, the mean is equal to the population mean, which is μ.

The standard deviation of the sampling distribution of sample means is given by

σ/√n.

The variance of a distribution is the square of the standard deviation; this means the variance of the sampling distribution of sample means would be

(σ/√n)² = σ²/(√n)² = σ²/n

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Need help ASAP!!!!!!
Sergeu [11.5K]

Input Data :

Point 1 ( x A , y A )  = (-6, -2)

Point 2 ( x B , y B )  = (1, -6)

Objective :

Find the distance between two given points on a line.

Formula :

Distance between two points =  √ ( x B − x A )2 + ( y B − y A ) 2

Solution :

Distance between two points =  √ ( 1 - -6 ) 2 + ( − 6 − − 2 ) 2

=  √ 7 2 + ( − 4 ) 2

=  √ 49 + 16

= √ 65  = 8.0623

Distance between points (-6, -2) and (1, -6) is 8.0623

Or 8!!

Best of luck!!

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