Individual taxi fares in Metro City are normally distributed with a population mean equals to $10 and a population standard deviation of $4. Assume drawing from an infinite population, a random sample of 64 fares is obtained. Based on these information, what is the probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10

Answer:

0.3413 = 34.13% probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10.

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 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 [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.

Mean equals to $10 and a population standard deviation of $4

This means that [tex]\mu = 10, \sigma = 4[/tex]

Sample of 64

This means that [tex]n = 8, s = \frac{4}{\sqrt{64}} = 0.5[/tex]

What is the probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10?

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 9.5.

X = 10

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{10 - 10}{0.5}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.5

X = 9.5

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{9.5 - 10}{0.5}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a pvalue of 0.1587

0.5 - 0.1587 = 0.3413

0.3413 = 34.13% probability that the sample mean fare of the 64 fares in the sample is between $9.5 and $10.

Exercise 4
3
2.5% = 50
2 * = 0,125
Solve the following equations:
1 3* = 81
2
5Y - 15 = 0
5
5x + 1 = 1
16* = 8
9.
1
9
4
8.
101 - x = 0,01
2(5)* = 54
(3)
5Y - 0,2 = 0
12
7 (-3)* = 1
11
5-2 = 25
10
107
Unit 3: Calculations with exponents​

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