Question

In a factory, the average number of chairs manufactured per day is 60, and the standard deviation is 5. If a day is selected at random, find these probabilities. SHow your work:

a. at least 68 chairs will be manufactured.

b. at most 57 chairs will be manufactured.

c. between 54 and 73 chairs will be manufactured.

Answer 1

Solution :

Given that,

mean = = 60

standard deviation = =5

P(x > 68) = 1 - P(x< 68)

= 1 - P[(x -) / < (68-60) /5 ]

= 1 - P(z <1.6 )

Using z table

= 1 -  0.9452

probability= 0.0548

b.

P(x< 57)

=P[(x -) / < (57-60) /5 ]

= P(z <-0.6 )

Using z table

=0.2743

probability= 0.2743

c

P(54< x < 73) = P[(54-60) /5 < (x - ) / < (73-60) /5]

= P(-1.2 < Z <0.24 )

= P(Z < 0.24) - P(Z < -1.2)

Using z table   

=0.5948 -   0.1151

probability= 0.4797