Math Help Please! (Show your work)
1 answer:
You might be interested in
Using the <u>normal distribution and the central limit theorem</u>, it is found that the interval that contains 99.44% of the sample means for male students is (3.4, 3.6).
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
In this problem:
- The mean is of
.
- The standard deviation is of
.
- Sample of 100, hence
The interval that contains 95.44% of the sample means for male students is <u>between Z = -2 and Z = 2</u>, as the subtraction of their p-values is 0.9544, hence:
Z = -2:
By the Central Limit Theorem
Z = 2:
The interval that contains 99.44% of the sample means for male students is (3.4, 3.6).
You can learn more about the <u>normal distribution and the central limit theorem</u> at brainly.com/question/24663213
Answer:
(x,y)=(0,7.333)
Step-by-step explanation:
We are required to:
Maximize p = x + 2y subject to
- x + 3y ≤ 22
- 2x + y ≤ 14
- x ≥ 0, y ≥ 0.
The graph of the lines are plotted and attached below.
From the graph, the vertices of the feasible region are:
- (0,7.333)
- (4,6)
- (0,0)
- (7,0)
At (0,7.333), p=0+2(7.333)=14.666
At (4,6), p=4+2(6)=4+12=16
At (0,0), p=0
At (7,0), p=7+2(0)=7
Since 14.666 is the highest, the maximum point of the feasible region is (0,7.333).
At x=0 and y=7.333, the function p is maximized.
