Help me stop airing I pay u £5

Norma-Jean [14]

What is the question supposed to be?

5 0

3 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

IgorLugansk [536]

Answer:

(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.85

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 20

$\mu$ = true average porosity

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.85-1.96 \times {\frac{0.75}{\sqrt{20} } }$ , $4.85+1.96 \times {\frac{0.75}{\sqrt{20} } }$ ]

= [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.56

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 16

$\mu$ = true average porosity

Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 98% confidence interval for the true mean, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1% level

of significance are -2.3263 & 2.3263}

P(-2.3263 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.3263$ ) = 0.98

P( $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } }$ , $4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } }$ ]

= [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

7 0

3 years ago

Simplify the expression: 3x + 2x + 10

PSYCHO15rus [73]

Answer:

5x+10

Step-by-step explanation:

3x+2x=5x

5x +10

3 0

3 years ago

Read 2 more answers

Question below here you can rear if not aks me please and thank you

Vanyuwa [196]

180-48.2-75=56.8 thank you

5 0

3 years ago

Start with the basic function f(x)=2x. If you have an initial value of 1, then you end up with the following iterations. f(1)=2*

denpristay [2]

f(x)=2x

f(1)=2*1=2

f^2(1)=2*2*1=4

f^3(1) =2*2*2*1=8

1. If you continue this pattern, what do you expect would happen to the numbers as the number of iterations grows?

I expect the numbers continue growing multiplying each time by 2.

Check your result by conducting at least 10 iterations.

f^4(1) = f^3(1) * f(1) = 8*2 = 16

f^(5)(1) = f^4(1) * f(1) = 16 * 2 = 32

f^6 (1) = f^5 (1) * f(1) = 32 * 2 = 64

f^7 (1) = f^6 (1) * f(1) = 64 * 2 = 128

f^8 (1) = f^7 (1) * f(1) = 128 * 2 = 256

f^9 (1) = f^8 (1) * f(1) = 256 * 2 = 512

f^10 (1) = f^9 (1) * f(1) = 512 * 2 = 1024

2. Repeat the process with an initial value of −1. What happens as the number of iterations grows?

f(-1) = 2(-1) = - 2

f^2 (-1) = f(-1) * f(-1) = - 2 * - 2 = 4

f^3 (-1) = f^2 (-1) * f(-1) = 4 * (-2) = - 8

f^4 (-1) = f^3 (-1) * f(-1) = - 8 * (-2) = 16

f^5 (-1) = f^4 (-1) * f(-1) = 16 * (-2) = - 32

As you see the magnitude of the number increases, being multiplied by 2 each time, and the sign is aleternated, negative positive negative positive ...

4 0

3 years ago