Answer:
The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.
Step-by-step explanation:
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 proportions, the mean is
and the standard deviation is ![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)
In this problem, we have that:
.
So
![\mu = 0.82](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.82)
![s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.82*0.18}{225}} = 0.0256](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%7B0.82%2A0.18%7D%7B225%7D%7D%20%3D%200.0256)
The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.
Answer:
b= -4
Step-by-step explanation:
On both sides of the equation has distributive property. We need to solve that first. Let's start with the left side of the equation.
*Positive multiplied by a negative equals a negative
-2b+2(b-10)= 2(10+5b)
-2b+2b-20= 2(10+5b)
-20= 2(10+5b)
Now let's solve the right side of the equation
-20= 20+10b
-20 -20
-40 = 10b
-40/10 = 10b/10
b= -4
Hope this helps!
£1=100p
100-22=78p
He gets 78p