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Sagot :
Answer:
0.0118 = 1.18% probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%
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 establishes 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.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
A telephone exchange operator assumes that 7% of the phone calls are wrong numbers.
This means that [tex]p = 0.07[/tex]
Sample of 459 phone calls:
This means that [tex]n = 459[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.07[/tex]
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\sqrt{\frac{0.07*0.93}{459}}} = 0.0119[/tex]
What is the probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%?
Proportion below 0.07 - 0.03 = 0.04 or above 0.07 + 0.03 = 0.1. Since the normal distribution is symmetric, these probabilities are the same, which means that we find one of them and multiply by 2.
Probability the proportion is below 0.04.
p-value of Z when X = 0.04. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.04 - 0.07}{0.0119}[/tex]
[tex]Z = -2.52[/tex]
[tex]Z = -2.52[/tex] has a p-value of 0.0059
2*0.0059 = 0.0118
0.0118 = 1.18% probability that the proportion of wrong numbers in a sample of 459 phone calls would differ from the population proportion by more than 3%
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