Answer:
The probability that the mean price for the sample was between $2.6982.698 and $2.7172.717 that week is P=0.726.
Step-by-step explanation:
We have a population mean price of $2.704. The standard deviation is $0.044.
We draw a sample of size n=31 out of this population.
We have to calculate the probability that the mean price for this sample is between $2.698 and $2.717 that week.
To calculate this, first we calculate the z-scores for both values:
X1=2.698
[tex]z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{2.698-2.704}{0.044/\sqrt{31}}=\dfrac{-0.006}{0.0079}=-0.7592[/tex]
X2=2.717
[tex]z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{2.717-2.704}{0.044/\sqrt{31}}=\dfrac{0.013}{0.0079}=1.645[/tex]
Then, we have:
[tex]P=P(2.698<x<2.717)=P(-0.7592<z<1.645)\\\\P=P(z<1.645)-P(z<-0.7592)\\\\P=0.95002-0.22387=0.72615[/tex]
