Answer:

And  and if we use the following function on the Ti84 plus we got:
 and if we use the following function on the Ti84 plus we got:
invNorm(0.02,0,1)
invNorm(1-0.02,0,1)
And the values with the middle 96% of the values are:

Step-by-step explanation:
For this case we want to find the limits with the middle 96% of the area below the normal curve, then the significance level would be:

And  and if we use the following function on the Ti84 plus we got:
 and if we use the following function on the Ti84 plus we got:
invNorm(0.02,0,1)
invNorm(1-0.02,0,1)
And the values with the middle 96% of the values are:

 
        
             
        
        
        
<span>6/3+(-1/3)
</span><span>= 6/3-1/3
= 5/3
= 1 2/3</span>
        
             
        
        
        
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
 and standard deviation  , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean  and standard deviation
 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
 and the standard deviation is 
In this problem, we have that:
 .
.
So


The mean of the sampling distribution of the sample proportions is 0.82 and the standard deviation is 0.0256.
 
        
                    
             
        
        
        
There is nothing in the space. If there were, then the space would not be blank.
        
             
        
        
        
Answer:70
Step-by-step explanation: