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:
$6 = cost of small box
$8 = cost of large box
Step-by-step explanation:
Let s = cost of small box
l = cost of large box
(1) 12s + 3l = 96 (2) 6s + 6l = 84
Multiply by -2 <u> -24s - 6l = -192</u>
-18s = - 108
s = $6 = cost of small box
12(6) + 3l = 96
72 + 3l = 96
3l = 24
l = $8 = cost of large box
Answer: 3424.28
Step-by-step explanation:
Given data : 378, 361, 350, 375, 200, 391, 375, 368, 321
Number of data values = 9
Mean :

Variance = 

Variance = 
Answer:
everything can be found in the picture above. I hope it helps
What do you mean by this? Do you have a picture