Answer:
<u>The sample system is given below:</u>
<em>see the attached for the graph</em>
The points (1, 2) and (4, -3) are in the intersected region but the point (-2, 8) is outside
The<u> profit</u> genereted by the <u>sales of gold label and black label combined</u> is <u>the sum</u> of the profits generated by the sales of gold label and of the profits generated by the sales of black label.
1. From the table
you can find the profits generated by the sales of gold label:
2. From the table
you can find the profits generated by the sales of black label:
In total
Answer: correct option 5
Answer:
a) 0.0392
b) 0.4688
c) At least $39,070 to be among the 5% most expensive.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?
This is the pvalue of Z when X = 20000. So
has a pvalue of 0.0392.
So this probability is 0.0392.
b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?
This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.
X = 30000
has a pvalue of 0.5080.
X = 20000
has a pvalue of 0.0392.
So this probability is 0.5080 - 0.0392 = 0.4688
c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?
This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.
The wedding would have to cost at least $39,070 to be among the 5% most expensive.