Answer:
The sampling distribution of sample proportion is approximately normal, with mean 0.62 and standard deviation 0.0485.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
62% of those people get some relief from taking ibuprofen (true proportion).
This means that ![p = 0.62](https://tex.z-dn.net/?f=p%20%3D%200.62)
Sample of 100
This means that ![n = 100](https://tex.z-dn.net/?f=n%20%3D%20100)
A. (4 pts.) Determine the sampling distribution of sample proportion. Also, find the mean and standard deviation of the sampling distribution.
By the Central Limit Theorem, it is approximately normal with
Mean ![\mu = p = 0.62](https://tex.z-dn.net/?f=%5Cmu%20%3D%20p%20%3D%200.62)
Standard deviation ![s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.62*0.38}{100}} = 0.0485](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.62%2A0.38%7D%7B100%7D%7D%20%3D%200.0485)
The option c) x < -24 represents the solution set of the given inequality.
<u>Step-by-step explanation</u>:
The given inequality is −1/4x−12 > −6
<u>To find the solution set :</u>
Keep the x term on one side and constants on other side of the inequality.
-1/4x > -6+12
-1/4x > 6
-x > (6
4)
-x > 24
Multiply both sides with -1 to remove the negative sign of x.
While multiplying or dividing a negative number in an equality, the inequality sign must be flipped.
⇒ x < -24
What are the answers for the nonshelaunt question
Answer:
Almost 125 bats (124.92) can be lined up from home plate to the center field fence.
Step-by-step explanation:
We have a baseball field was measured to be 125 yards from Home Plate to deep center field.
We have to calculate how many baseball bats (91.5 cm in length) could be lined up (end to end) from home plate to the center field fence.
One yard is equivalent to 0.9144 meters.
One meter is equivalent to 100 cm.
Let D the distance from home plate to the center field fence and L the bat length.
Then, the amount of bats b that can be lined up is:
![b=\dfrac{D}{L}=\left(\dfrac{125\;\text{yd}}{91.5\;\text{cm}}\right)\cdot \left(\dfrac{0.9144\;\text{m}}{1\;\text{yd}}\right)\cdot \left(\dfrac{100\;\text{cm}}{1\;\text{m}}\right)\\\\\\b=\dfrac{(125)(0.9144)(100)}{91.5}=\dfrac{11,430}{91.5}=124.92](https://tex.z-dn.net/?f=b%3D%5Cdfrac%7BD%7D%7BL%7D%3D%5Cleft%28%5Cdfrac%7B125%5C%3B%5Ctext%7Byd%7D%7D%7B91.5%5C%3B%5Ctext%7Bcm%7D%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cdfrac%7B0.9144%5C%3B%5Ctext%7Bm%7D%7D%7B1%5C%3B%5Ctext%7Byd%7D%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cdfrac%7B100%5C%3B%5Ctext%7Bcm%7D%7D%7B1%5C%3B%5Ctext%7Bm%7D%7D%5Cright%29%5C%5C%5C%5C%5C%5Cb%3D%5Cdfrac%7B%28125%29%280.9144%29%28100%29%7D%7B91.5%7D%3D%5Cdfrac%7B11%2C430%7D%7B91.5%7D%3D124.92)