Question

# Find the probability and interpret the results. If​ convenient, use technology to find the probability. During...

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 32 gas stations is drawn from this population. What is the probability that the mean price for the sample was between ​\$2.69 and ​\$2.72 that​ week? Assume σ equals=​\$0.047. The probability that the sample mean was between \$2.696 and ​\$2.728 is

#### Homework Answers

Answer #1

given mean=2.717

Standard Deviation=0.047

n=32

Here we take the distribution as normal distribution. Hence,

p(z1<z<z2)=p((2.69-2.717)/(0.047/sqrt(32))<(x-2.717)/(0.047/sqrt(32))<(2.72-2.717)/(0.047/sqrt(32))

=p(-3.25<z<0.36

=p(z=0.36)-p(z=-3.25)

=0.5254-0.2828

=0.2426.

p(z1<z<z2)=p((2.696-2.717)/(0.047/sqrt(32))<(x-2.717)/(0.047/sqrt(32))<(2.728-2.717)/(0.047/sqrt(32))

=p(-2.53<z<1.32)

=p(z=1.32)-p(z=-2.53)

=0.5925-0.3275

=0.265

=0.2426.

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