Respuesta :
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%.
Step-by-step 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu*n[/tex] and the standard deviation is [tex]s = \sigma \sqrt{n}[/tex]
In this problem, we have that:
[tex]\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160[/tex]
This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.
X = 4000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{4000 - 3500}{160}[/tex]
[tex]Z = 3.13[/tex]
[tex]Z = 3.13[/tex] has a pvalue of 0.9991
X = 3000
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3000 - 3500}{160}[/tex]
[tex]Z = -3.13[/tex]
[tex]Z = -3.13[/tex] 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%.
