In: Statistics and Probability
Salaries for teachers in a particular elementary school district have a mean of $44,000 and a standard deviation of $6,500. We randomly survey 36 teachers from that district.
Solution :
Given that ,
mean = = 44000
standard deviation = = 6500
n = 36
= 44000
= / n = 6500 / 36 = 1083.3333
The sampling distribution of mean = 44000
The standard deviation of the sampling mean = 1083.3333
P( < 43000 ) = P(( - ) / < ( 43000 - 44000 ) / 1083.3333 )
= P(z < -0.92)
Using z table
= 0.1788
Probability = 0.1788
P( 45000 < < 47000 )
= P[( 45000 - 44000) / < ( - ) / < ( 47000 - 44000) / 1083.3333 )]
= P(0.92 < Z < 2.77 )
= P(Z < 2.77 ) - P(Z < 0.92 )
Using z table,
= 0.9972 - 0.8212
= 0.1760
Probability = 0.1760