Answer:
The proportion of children that have an index of at least 110 is 0.0478.
Step-by-step explanation:
The given distribution has a mean of 90 and a standard deviation of 12.
Therefore mean, 
= 90 and standard deviation, 
= 12.
It is given to find the proportion of children having an index of at least 110.
We can take the variable to be analysed to be x = 110.
Therefore we have to find p(x < 110), which is left tailed.
Using the formula for z which is p( Z < 
) we get p(Z < 
= 1.67).
So we have to find p(Z ≥ 1.67) = 1 - p(Z < 1.67)
Using the Z - table we can calculate p(Z < 1.67) = 0.9522.
Therefore p(Z ≥ 1.67) = 1 - 0.9522 = 0.0478
Therefore the proportion of children that have an index of at least 110 is 0.0478
Your answer would be -624 because 1-210-3= -212 then 3 times -212 +12= -624
Answer:
The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
For this problem, we have that:
90% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).
The <em><u>correct answer</u></em> is:
$72.31 – $24.61 – $16.49; $31.21
Explanation:
We know we end the month with $72.31.
We made two deposits during the month and no other activity. This means if we take the amounts of the deposits away from the total at the end of the month, we can find how much we had at the beginning of the month:
72.31-24.61-16.49 = 31.21
Okay let's see.. 0.750
1.200
As we see 1 is bigger than 0.So that means 1/2 is bigger.Hope this helps! (;
