Answer:
To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.
Explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For sums, we have that the mean is
and the standard deviation is
![s = \sigma √(n)](https://img.qammunity.org/2021/formulas/mathematics/college/6gu6jhg73lbxus5f7f6a9ytx38h2j9mue3.png)
In this problem, we have that:
![\mu = 100*35 = 3500, \sigma = √(100)*16 = 160](https://img.qammunity.org/2021/formulas/mathematics/college/y73s7dv17jahetzbankcu519gsmynr1xxs.png)
This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.
X = 4000
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
By the Central Limit Theorem
![Z = (X - \mu)/(s)](https://img.qammunity.org/2021/formulas/mathematics/college/qbjdi63swemoz9mdzfqtue91aagng8mdqs.png)
![Z = (4000 - 3500)/(160)](https://img.qammunity.org/2021/formulas/mathematics/college/expmatujjrz6zxlhfxwkt0zp1sgayb6wox.png)
![Z = 3.13](https://img.qammunity.org/2021/formulas/mathematics/college/54q4r14mv49laq0d9p3a2caymh21kzqskm.png)
has a pvalue of 0.9991
X = 3000
![Z = (X - \mu)/(s)](https://img.qammunity.org/2021/formulas/mathematics/college/qbjdi63swemoz9mdzfqtue91aagng8mdqs.png)
![Z = (3000 - 3500)/(160)](https://img.qammunity.org/2021/formulas/mathematics/college/r90grhz291gclv0icxu7cy9trog6pfsxvy.png)
![Z = -3.13](https://img.qammunity.org/2021/formulas/mathematics/college/de1svwr3746sfa8xuy6iwluda48f0n6uh5.png)
has a pvalue of 0.0009
0.9991 - 0.0009 = 0.9982
So the correct answer is:
To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.