find the probability and interpret the results if convenient

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 $2.712 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 $2.695 and $2.729 that week? assume o=$0.049.

a) the probability that the sample mean was between $2.695 and $2.729 is ???

Solution

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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 $2.717 per gallon. A random sample of 31 gas stations is drawn from this population. What is the probability that the mean price for the sample was between $2.694 and$2.718 that week? Assume ? = $0.049.

Answer

\"Normal Distribution

Mean ( u ) =2.717

Standard Devaition ( sd )=0.049

Number ( n ) = 31

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

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

P(X < 2.694) = (2.694-2.717)/0.049/ Sqrt ( 31 )

= -0.023/0.0088

= -2.6134

= P ( Z <-2.6134) From Standard Normal Table

= 0.00448

P(X < 2.718) = (2.718-2.717)/0.049/ Sqrt ( 31 )

= 0.001/0.0088 = 0.1136

= P ( Z <0.1136) From Standard Normal Table

= 0.54523

P(2.694 < X < 2.718) = 0.54523-0.00448 = 0.5408