Respuesta :
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 [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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], 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]
In this problem, we have that:
[tex]\mu = 2, \sigma = 0.05, n = 25, s = \frac{0.05}{\sqrt{25}} = 0.01[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2 - 2}{0.01}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5
X = 1.99
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1.99 - 2}{0.01}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] 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
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1.98 - 2}{0.01}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] 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
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2.01 - 2}{0.01}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] 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.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]2.325 = \frac{X - 2}{0.01}[/tex]
[tex]X - 2 = 0.01*2.325[/tex]
[tex]X = 2.02325[/tex]
