Question

please show work

3. Students taking Professor’s Angela Mazza’s Introduction to Marketing course spent an average of 1.5 hours to complete an assignment with a standard deviation of 0.40 hours and it follows the normal probability distribution.

  

(a) Find the portion of the students who spent between 1.5 and 2.5 hours to complete an assignment.

(b) Find the portion of the students who spent more than 2.5 hours to complete an assignment.

(c) Find the portion of the students who spent between 2.5 and 2.7 hours to complete an assignment.

(d) Find the portion of the students who spent between 1 and 2.7 hours to complete an assignment.

Answer 1

Solution :

Given that ,

mean = = 1.5

standard deviation = = 0.40

P(1.5 < x <2.5 ) = P[( 1.5 - 1.5 ) / 0.40 ) < (x - ) /  < ( 2.5 - 1.5 ) / 0.40 ) ]

= P( 0 < z < 2.5 )

= P(z < 2.5 ) - P(z < 0 )

Using z table,

= 0.9938 - 0.5

= 0.4938

P(1.5 < x <2.5 ) = 0.4938

P(x >2.5 ) = 1 - P( x < 2.5 )

=1- P[(x - ) / < ( 2.5 - 1.5 ) / 0.40 ]

=1- P(z < 2.5 )

Using z table,

= 1 - 0.9938

= 0.0062

P(x >2.5 ) = 0.0062

P( 2.5 < x < 2.7 )

= P[( 2.5 - 1.5 ) / 0.40 ) < (x - ) /  < ( 2.7 - 1.5 ) / 0.40) ]

= P( 2.5 < z < 3 )

= P(z < 3 ) - P(z < 2.5 )

Using z table,

= 0.9987 - 0.9938

= 0.0049

P( 2.5 < x < 2.7 ) = 0.0049

P( 1 < x < 2.7 )

= P[( 1 - 1.5 ) / 0.40 ) < (x - ) /  < ( 2.7 - 1.5 ) / 0.40) ]

= P( -1.25 < z < 3)

= P(z < 3 ) - P(z < -1.25 )

Using z table,

= 0.9987 - 0.1056

= 0.8931

P( 1 < x < 2.7 ) = 0.8931