Question 1177434: The amount of electricity bills for all households in a city have screwed probability distribution with a mean of $140 and a standard deviation of $30. Find the probability that the mean amount of electricity bills for a random sample of 75 selected from this city will be more than the population mean by at least 4.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step by step.
1. **Define the Random Variable**
Let X be the random variable representing the amount of electricity bills for all households in the city. We are given that X has a skewed distribution with a mean (µ) of $140 and a standard deviation (σ) of $30.
2. **Central Limit Theorem**
Since we are dealing with a sample mean (x̄) from a skewed distribution, we can apply the Central Limit Theorem (CLT). The CLT states that the distribution of sample means will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is large enough (typically n ≥ 30). In this case, we have a sample size of 75, which is sufficient to apply the CLT.
3. **Calculate the z-score**
We want to find the probability that the sample mean (x̄) is greater than the population mean (µ) by at least $4. This can be expressed as:
P(x̄ ≥ µ + 4) = P(x̄ ≥ 140 + 4) = P(x̄ ≥ 144)
To standardize this, we calculate the z-score:
z = (x̄ - µ) / (σ / √n) = (144 - 140) / (30 / √75) ≈ 1.15
4. **Find the Probability**
Now, we need to find the probability that the z-score is greater than 1.15. Using a standard normal table or calculator, we find:
P(Z ≥ 1.15) ≈ 0.1251
Therefore, the probability that the mean amount of electricity bills for a random sample of 75 households will be more than the population mean by at least $4 is approximately **0.1251**.
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