Find the probability and interpret the results. If convenient, use technology to

Question

Find the probability and interpret the results. If convenient, use technology to find the probability. During a certain week the mean price of gasoline was $2.709 per gallon. A random sample of 33 gas stations is drawn from this population. What is the probability that the mean price for the sample was between $2.689 and $2.728 that week? Assume o=$0.05

The Probability that the sample mean was between $2.689 and $2.728 is ____

Interpret the results. Choose the correct answer below:

A. About 3% of samples of 33 gas stations that week will have a mean price between $2.689 and $2.728.

B.About 97% of samples of 33 gas stations that week will have a mean price between $2.689 and $2.728.

C. About 3% of the population of 33 gas stations that week will have a mean price between $2.689 and $2.728.

D.About 97% of the population of 33 gas stations that week will have a mean price between $2.689 and $2.728.

Explanation / Answer

The correct option is B

Explanantion

Normal Distribution

Mean ( u ) =2.709

Standard Devaition ( sd )=0.05

Number ( n ) = 33

Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)"

To find P(a <= Z <=b) = F(b) - F(a)

P(X < 2.689) = (2.689-2.709)/0.05/ Sqrt ( 33 )

= -0.02/0.0087

= -2.2988

= P ( Z <-2.2988) From Standard Normal Table

=1P ( Z<2.2988 )

=10.9893

=0.0107

P(X < 2.728) = (2.728-2.709)/0.05/ Sqrt ( 33 )

= 0.0019/0.0087 = 2.1839

= P ( Z <2.1839) From Standard Normal Table

= 0.9854

P(2.694 < X < 2.718) = 0.9854 - 0.0107= 0.9747

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