Furnace repair bills are normally distributed with a mean of 267 dollars and a standard deviation of 20 dollars. If 64 of these repair bills are randomly selected, find the probability that they have a mean cost between 267 dollars and 269 dollars.

Respuesta :

Answer: 0.7881446

Step-by-step explanation:

Given : Mean : [tex]\mu = 267\text{ dollars} [/tex]

Standard deviation : [tex]\sigma =20 \text{ dollars}[/tex]

Sample size : [tex]n=64[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x=  267 dollars

[tex]z=\dfrac{267-267}{\dfrac{20}{\sqrt{64}}}=0[/tex]

For x= 269 dollars.

[tex]z=\dfrac{269-267}{\dfrac{20}{\sqrt{64}}}=0.80[/tex]

The P-value : [tex]P(0<z<0.8)=P(z<0.8)-P(z<0)[/tex]

[tex]= 0.7881446-0.5= 0.2881446\approx 0.7881446[/tex]

Hence, the probability that they have a mean cost between 267 dollars and 269 dollars.= 0.7881446

Answer:

There is a 28.81% probability that they have a mean cost between 267 dollars and 269 dollars.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 267, \sigma = 20, n = 64, s = \frac{20}{\sqrt{64}} = 2.5[/tex].

Find the probability that they have a mean cost between 267 dollars and 269 dollars.

This probability is the pvalue of Z when X = 269 subtracted by the pvalue of Z when X = 267. So:

X = 269

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

[tex]Z = \frac{269 - 267}{2.5}[/tex]

[tex]Z = 0.8[/tex]

[tex]Z = 0.8[/tex] has a pvalue of 0.7881.

X = 267

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

[tex]Z = \frac{267 - 267}{2.5}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.50.

So there is a 0.7881 - 0.50 = 0.2881 = 28.81% probability that they have a mean cost between 267 dollars and 269 dollars.