Question

A simple random sample of size n=36 is obtained from a population with mean=89 and standard deviation = 6

​(c) What is P ( x overbar less than or equal to 86.65)​? ​

(d) What is P(88.5 < x overbar < 91.25)?

​(a) Describe the sampling distribution of x overbar.

​(b) What is P ( x overbar > 90.85 )?

Answer 1

Solution :

Given that ,

mean = = 89

standard deviation = = 6

n = 36

= 89

=  / n = 6 / 36=1

(A)P( < 86.65) = P[( - ) / < (86.65 - 89) / 1]

= P(z <-2.35 )

Using z table

=0.0094

(B)P(88.5<    <91.25 ) = P[(88.5 - 89) /1 < ( - ) / < (91.25 - 89) /1 )]

= P( -0.5< Z <2.25 )

= P(Z <2.25 ) - P(Z <-0.5 )

Using z table,  

= 0.9878 - 0.3085

= 0.6793

(C)P( > 90.85) = 1 - P( <90.85 )

= 1 - P[( - ) / < (90.85 - 89) / 1]

= 1 - P(z <1.85 )

Using z table,    

= 1 - 0.9678

=0.0322