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