Answer:
The answer to the question is;
The confidence level decreases.
Step-by-step explanation:
The magnitude of confidence level reflects the utility level of how we estimate the population parameter. That is an indication of the confidence level specifies the utility level of related statistical parameter.
The magnitude of the confidence level is inversely proportional to the utility level. That is an increase in the confidence level results in a decrease in the utility level and vise versa.
For example, if the confidence level that there will be a scarcity in an essential commodity in a few days is 99 % then shoppers would divert their resoucrces to purchase that essential commodity, that is the utility of the money shoppers have is reduced as it is mostly focused on the essential commodity.
If the confidence level that there would be a scarcity was stated as 95 % then the money shoppers have to spend will be diverted to a wider variety of items, therefore, rising the utility of the money.
If Jerry contributes at the beginning of the month and withdraws at the end of the month, the final contribution earns 1 month's interest. The one before that earns 2 months' interest, so has a value of (1+0.017/12) times that of the last payment. In short, the sum is that of a geometric sequence with first term
a₁ = 300*(1+0.017/12)
and common ratio
r = 1+0.017/12
We assume Jerry contributes each month for 15 years, so a total of 180 payments. The sum is given by the formula for the sum of a geometric sequence.
Filling in your numbers, this is
If Jerry's contributions and withdrawal are at the end of the month, this balance is reduced by 1 month's interest, so is $61,460.
_____
We suppose the expected choice is $61,960. This supposition comes from the fact that a handwritten 4 is often confused with a handwritten 9. The usual simple calculation of future value uses end-of-the-month contributions by default. (a₁ = 300)
Answer:
P(45<X<81) = P(-3<z<3)=0.9974 or 99.74%
Step-by-step explanation:
Mean = 63
Standard Deviation =6
We need to find probability that X lies between 45 and 81 i.e P(45<X<81)
First we will find z-score for X=45 and X=81
For X=45 the z-score can be found using formula:
Putting values and finding z-score
For X=45, z-score is -3
For X=81 the z-score can be found using formula:
Putting values and finding z-score
For X=81, z-score is 3
Now finding the probability of P(-3<z<3)
It can be solved as:
P(-3<z<3)=P(z<3)-P(z<-3)
Looking at the z-score table to find
P(z<3)= 0.9987
P(z<-3)=0.0013
So, P(-3<z<3)=P(z<3)-P(z<-3)
P(-3<z<3)=0.9987 - 0.0013
P(-3<z<3)=0.9974
or P(-3<z<3)= 99.74%
So, P(45<X<81) = P(-3<z<3)=0.9974 or 99.74%
Answer:5/2 times 1/3
Step-by-step explanation:
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Answer:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).
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 .
Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.
This means that
80% 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 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).