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Pachacha [2.7K]
3 years ago
10

g PROBLEM 3. Based on data from Consumer Reports, replacement times of TVs is on average 3.75 years with the standard deviation

of 2.5 years. Answer the following questions. Question 1 (3 points). A random sample of 28 TVs is chosen. What is the probability that their average replacement time is less than 3.5 years
Mathematics
1 answer:
Aleksandr [31]3 years ago
4 0

Answer:

29.81% probability that their average replacement time is less than 3.5 years

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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 3.75, \sigma = 2.5, n = 28, s = \frac{2.5}{\sqrt{28}} = 0.47

What is the probability that their average replacement time is less than 3.5 years

This is the pvalue of Z when X = 3.5. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{3.5 - 3.75}{0.45}

Z = -0.53

Z = -0.53 has a pvalue of 0.2981

29.81% probability that their average replacement time is less than 3.5 years

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