Answer:
a) 0.1587
b) 0.0008
Step-by-step explanation:
Given that:
Mean (m) = 37 cm
Sample size (n) = 10
standard deviation (s) = 5 cm
a) the z score (z) = [tex]\frac{x-m}{s}[/tex]
where x = 42 cm
Therefore, [tex]z=\frac{x-m}{s}=\frac{42-37}{5} =1[/tex]
The probability that all 10 have a calf circumference greater than 42 cm =
P(x > 42) = P (z > 1) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587
b) The standard error of the mean is a measure of how far that the sample mean will be from the population mean for a repeated random samples of size n.
The standard error of the mean (e)= s /√n
e = s /(√n ) = 5 / (√10) = 1.5811
The z score (z) = [tex]\frac{x-m}{e}[/tex]
where x = 42 cm
Therefore, [tex]z=\frac{x-m}{e}=\frac{42-37}{1.5811} = 3.16[/tex]
P(x > 42) = P (z > 3.16) = 1 - P(z < 3.16) = 1 - 0.9992 = 0.008
