3 part question. If answer all(brainilest will be given.

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

Invasive species often experience exponential population growth when introduced into a new environment. Zebra mussels are an inv

Oliga [24]

Answer:

If ‘a’ is the initial population of the Zebra mussels, then every six months, the population of Zebra mussels quadruples, i.e. it becomes 4a. In t years, i.e. in 2t intervals of 6 months each, the population of Zebra mussels can be computed with the help of a geometric series a, ar, ar2, ar3, …. arn-1 ( the series being finite with n terms) where a is the 1st term , r is the common ratio and n is the number of terms in the series..

Here, in 2t years, there will be 2t + 1 terms in the geometric series including the initial term. Thus in the above series, a = 10, r = 4 and n = 2t + 1. Then the population of the Zebra mussels after 2t years is the (2t +1)st term of the geometric series I.e. 10* ( 42t)

In 15 months, t = 15/12 = 1.25. Then the population of Zebra mussels after 15 months will be 10*(42.5 ) = 10 * 25 = 320

If after t years, the population of the Zebra mussels become 1 million , then we have

1000000 = 10 * (42t) or, 100000= 42t or,105 = 42t Taking logarithms of both sides, we have 5 log 10 = 2t log 4 or, t = (5log10)/(2log4) = 5/1.20 years or (5/1.20) * 12 months = 50 months, i.e. 4 years and 2 months.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Help (20 points, due tmr!! ) *pic below* :)

sveta [45]

Angles:
2 acute angles and 2 obtuse angles

Sides:
2 pairs pf parallel sides

7 0

3 years ago

What is the value of the following expression when d=2? 3d + (2.d)-4

Lostsunrise [7]

Answer: