Answer:
P(35.3 < M < 35.4) = 0.0040.
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
and standard deviation
, the zscore 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 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 random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 33.9, \sigma = 6.7, n = 119, s = \frac{6.7}{\sqrt{119}} = 0.6142](https://tex.z-dn.net/?f=%5Cmu%20%3D%2033.9%2C%20%5Csigma%20%3D%206.7%2C%20n%20%3D%20119%2C%20s%20%3D%20%5Cfrac%7B6.7%7D%7B%5Csqrt%7B119%7D%7D%20%3D%200.6142)
Find the probability that a single randomly selected value is between 35.3 and 35.4
This is the pvalue of Z when X = 35.4 subtracted by the pvalue of Z when X = 35.3. So
X = 35.4
![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{35.4 - 33.9}{0.6142}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B35.4%20-%2033.9%7D%7B0.6142%7D)
![Z = 2.44](https://tex.z-dn.net/?f=Z%20%3D%202.44)
has a pvalue of 0.9927
X = 35.3
![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{35.3 - 33.9}{0.6142}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B35.3%20-%2033.9%7D%7B0.6142%7D)
![Z = 2.28](https://tex.z-dn.net/?f=Z%20%3D%202.28)
has a pvalue of 0.9887
0.9927 - 0.9887 = 0.0040
So the answer is:
P(35.3 < M < 35.4) = 0.0040.