Answer:
a) P(X < 99) = 0.2033.
b) P(98 < X < 100) = 0.4525
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 normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 p-value, 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.
Mean of 100 and variance of 36.
This means that ![\mu = 100, \sigma = \sqrt{36} = 6](https://tex.z-dn.net/?f=%5Cmu%20%3D%20100%2C%20%5Csigma%20%3D%20%5Csqrt%7B36%7D%20%3D%206)
Sample of 25:
This means that ![n = 25, s = \frac{6}{\sqrt{25}} = 1.2](https://tex.z-dn.net/?f=n%20%3D%2025%2C%20s%20%3D%20%5Cfrac%7B6%7D%7B%5Csqrt%7B25%7D%7D%20%3D%201.2)
(a) P(X<99)
This is the pvalue of Z when X = 99. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{99 - 100}{1.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B99%20-%20100%7D%7B1.2%7D)
![Z = -0.83](https://tex.z-dn.net/?f=Z%20%3D%20-0.83)
has a pvalue of 0.2033. So
P(X < 99) = 0.2033.
b) P(98 < X < 100)
This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So
X = 100
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{100 - 100}{1.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B100%20-%20100%7D%7B1.2%7D)
![Z = 0](https://tex.z-dn.net/?f=Z%20%3D%200)
has a pvalue of 0.5
X = 98
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{98 - 100}{1.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B98%20-%20100%7D%7B1.2%7D)
![Z = -1.67](https://tex.z-dn.net/?f=Z%20%3D%20-1.67)
has a pvalue of 0.0475
0.5 - 0.0475 = 0.4525
So
P(98 < X < 100) = 0.4525