Answer: 0.1310
Step-by-step explanation:
Given : Mean : [tex]\mu = \text{131 millimeters}[/tex]
Standard deviation : [tex]\sigma = \text{7 millimeters}[/tex]
Sample size : [tex]n=31[/tex]
To find the probability that the sample mean would differ from the population mean by more than 1.9 millimeters i.e. less than 129.1 milliliters and less than 132.9 milliliters.
The formula for z-score :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x = 129.1 milliliters
[tex]z=\dfrac{129.1-131}{\dfrac{7}{\sqrt{31}}}\approx-1.51[/tex]
For x = 132.9 milliliters
[tex]z=\dfrac{132.9-131}{\dfrac{7}{\sqrt{31}}}\approx1.51[/tex]
The P-value= [tex]P(x<-1.51)+P(x>1.51)[/tex]
[tex]=2P(z>1.51)=2(1-P(z<1.15))\\=2(1-0.9344783)\\=0.1310434\approx0.1310[/tex]
Hence, the required probability = 0.1310
