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furnace repair bills are normally distributed with a mean of 273 dollar…

Question

furnace repair bills are normally distributed with a mean of 273 dollars and a standard deviation of 30 dollars. if 144 of these repair bills are randomly selected, find the probability that they have a mean cost between 273 dollars and 276 dollars. round to four decimal places

a. 0.2881
b. 0.7881
c. 0.2119
d. 0.5547

Explanation:

Step1: Calculate the standard error

The formula for the standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \), where \( \sigma = 30 \) (standard deviation) and \( n = 144 \) (sample size).

$$ \sigma_{\bar{x}}=\frac{30}{\sqrt{144}}=\frac{30}{12} = 2.5 $$

Step2: Calculate the z - scores

The formula for the z - score is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \).
For \( \bar{x}=273 \), \( z_1=\frac{273 - 273}{2.5}=0 \)
For \( \bar{x}=275 \), \( z_2=\frac{275 - 273}{2.5}=\frac{2}{2.5}=0.8 \)

Step3: Find the probability

We want to find \( P(0<Z<0.8) \). Using the standard normal distribution table, \( P(Z < 0.8)=0.7881 \) and \( P(Z < 0)=0.5 \)

$$ P(0

Answer:

A. 0.2881