Question

Assuming the density is not normal, with a mean of 5 and a standard deviation of 3.2, what is the probability that we will see between 10 and 20 birds of this kind on a random day?

What range of birds would have a probability of greater than 80%?

Answer 1

Solution :

Given that ,

mean = = 5

standard deviation = =3.2

P(10< x <20 ) = P[(10 -5) /3.2 < (x - ) / < (20 -5) /3.2 )]

= P( 1.5625< Z <4.6875 )

= P(Z <4.6875 ) - P(Z < 1.5625)

Using z table   

= 1-0.9409

   probability=0.0591

(B)

Using standard normal table,

P(Z > z) = 80%

= 1 - P(Z < z) = 0.80

= P(Z < z ) = 1 - 0.80

= P(Z < z ) = 0.20

= P(Z < - 0.84 ) = 0.20  

z =- 0.84 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= - 0.84 *3.2+5

x= 2.312

x=2