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Tomtit [17]
3 years ago
11

Suppose a simple random sample of size n = 14 is obtained from a population with μ = 61 and σ = 17.

Mathematics
1 answer:
sergejj [24]3 years ago
5 0

Answer:

C. The population must be normally distributed.

Step-by-step explanation:

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.

For us to apply the central limit theorem with a sample size of 14, the underlying population must be normally distributed.

So the correct answer is:

C. The population must be normally distributed.

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3 0
2 years ago
Set up, but do not evaluate, the integral that represents the length of the curve given by x = 1 + 3t^2, y = 4 + 2t^3 over the i
kherson [118]

L

=

∫

t

f

t

i

√

(

d

x

d

t

)

2

+

(

d

y

d

t

)

2

d

t

. Since  

x

and  

y

are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:

d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

And we substitute these into the integral:

L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

18

t

2

+

9

t

4

+

36

t

2

d

t

=

∫

√

3

0

√

9

+

18

t

2

+

9

t

4

d

t

=

∫

√

3

0

√

(

3

+

3

t

2

)

2

d

t

=

∫

√

3

0

(

3

+

3

t

2

)

d

t

=

3

t

+

t

3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=6The arclength of a parametric curve can be found using the formula:  

L

=

∫

t

f

t

i

√

(

d

x

d

t

)

2

+

(

d

y

d

t

)

2

d

t

. Since  

x

and  

y

are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:

d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

And we substitute these into the integral:

L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

18

t

2

+

9

t

4

+

36

t

2

d

t

=

∫

√

3

0

√

9

+

18

t

2

+

9

t

4

d

t

=

∫

√

3

0

√

(

3

+

3

t

2

)

2

d

t

=

∫

√

3

0

(

3

+

3

t

2

)

d

t

=

3

t

+

t

3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=

6

√

3

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.    

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.

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