Step-by-step explanation:
percentage of adults= number of adults over total number of persons times 100;
400/400+350*100
400/750*100
40/75*100
8/15*100
8/3*20
160/3
=53.3%
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:
C is false as the other three are 100% correct statements. I dont know what is linear growth and exponential growth but I know that the other 3 statements are correct so the only 1 left C must be false.
The weights are within 2 standard deviations of the mean are 8.9 lbs, 9.5 lbs and 10.4 lbs
<h3>How to determine the weights?</h3>
The given parameters are:
- Mean, μ = 9.5
- Standard deviation, σ = 0.5
The weights within 2 standard deviation is represented as:
μ - 2σ ≤ x ≤ μ + 2σ
Substitute known values
9.5 - 2(0.5) ≤ x ≤ 9.5 + 2(0.5)
Evaluate the product
9.5 - 1 ≤ x ≤ 9.5 + 1
Evaluate the sum
8.5 ≤ x ≤ 10.5
This means that the weights are between 8.5 and 10.5 (inclusive)
Hence, the weights are within 2 standard deviations of the mean are 8.9 lbs, 9.5 lbs and 10.4 lbs
Read more about standard deviation at:
brainly.com/question/11743583
No, 43 is a prime number, so that statement is false.