The length of zebra pregnancies is normally distributed, with mean μ = 380 and standard deviation...

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.

Solutions

Expert Solution

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

orchestra answered 1 year ago

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