In: Statistics and Probability
The length of zebra pregnancies is normally distributed, with mean μ = 380 and standard deviation σ = 10. A random sample of 11 random pregnant zebras is chosen. Find P(x⎯⎯⎯x¯ < 375) for n = 11.
Enter your answer as an area under the curve with 4 decimal places.
Solution :
Given that,
mean =
= 380
standard deviation =
= 10
n = 11
= 380
=
/
n = 10/
11=3.015
P(
<375 ) = P[(
-
) /
< (375-380) / 3.015]
= P(z < -1.66)
Using z table
=0.0485
probability= 0.0485