Answer:
a) 34.13% probability that the sample mean will be between 1.99 and 2 litres.
b) 2.28% probability that the sample mean will be below 1.98 litres.
c) 15.87% probability that the sample mean will be above 2.01 litres.
d) 2.02325 litres
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 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 = 2, \sigma = 0.05, n = 25, s = \frac{0.05}{\sqrt{25}} = 0.01](https://tex.z-dn.net/?f=%5Cmu%20%3D%202%2C%20%5Csigma%20%3D%200.05%2C%20n%20%3D%2025%2C%20s%20%3D%20%5Cfrac%7B0.05%7D%7B%5Csqrt%7B25%7D%7D%20%3D%200.01)
a. Between 1.99 and 2.0 litres?
This is the pvalue of Z when X = 2 subtracted by the pvalue of Z when X = 1.99. So
X = 2
![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{2 - 2}{0.01}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B2%20-%202%7D%7B0.01%7D)
![Z = 0](https://tex.z-dn.net/?f=Z%20%3D%200)
has a pvalue of 0.5
X = 1.99
![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{1.99 - 2}{0.01}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1.99%20-%202%7D%7B0.01%7D)
![Z = -1](https://tex.z-dn.net/?f=Z%20%3D%20-1)
has a pvalue of 0.1587
0.5 - 0.1587 = 0.3413
34.13% probability that the sample mean will be between 1.99 and 2 litres.
b. Below 1.98 litres?
pvalue of Z when X = 1.98. So
![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{1.98 - 2}{0.01}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1.98%20-%202%7D%7B0.01%7D)
![Z = -2](https://tex.z-dn.net/?f=Z%20%3D%20-2)
has a pvalue of 0.0228
2.28% probability that the sample mean will be below 1.98 litres.
c. Above 2.01 litres?
1 subtracted by the pvalue of Z when X = 2.01. So
![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{2.01 - 2}{0.01}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B2.01%20-%202%7D%7B0.01%7D)
![Z = 1](https://tex.z-dn.net/?f=Z%20%3D%201)
has a pvalue of 0.8413
1 - 0.8413 = 0.1587
15.87% probability that the sample mean will be above 2.01 litres.
d. The probability is 99% that the sample mean will contain at least how much soft drink?
Value of X when Z has a pvalue of 0.99. So X when Z = 2.325.
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![2.325 = \frac{X - 2}{0.01}](https://tex.z-dn.net/?f=2.325%20%3D%20%5Cfrac%7BX%20-%202%7D%7B0.01%7D)
![X - 2 = 0.01*2.325](https://tex.z-dn.net/?f=X%20-%202%20%3D%200.01%2A2.325)
![X = 2.02325](https://tex.z-dn.net/?f=X%20%3D%202.02325)