Question

# A machine discharges an average of 250 milliliters per cup with standard deviation of 17 milliliters....

A machine discharges an average of 250 milliliters per cup with standard deviation of 17 milliliters.

a) What percentage of cups will contain more than 260 milliliters?

b) Between what volumes do the middle 75% of cups fall?

c) If the cups have a maximum capacity of 275 milliliters, approximately how many of the next 1000 cups filled will overflow.

Solution :

Given that ,

mean = = 250

standard deviation = = 17

a) P(x > 260 ) = 1 - p( x< 260)

=1- p P[(x - ) / < (260 - 250) / 17]

=1- P(z < 0.59)

= 1 - 0.7224

= 0.2776

percent = 27.76%

b) Using standard normal table,

P( -z < Z < z) = 75%

= P(Z < z) - P(Z <-z ) = 0.75

= 2P(Z < z) - 1 = 0.75

= 2P(Z < z) = 1 + 0.75

= P(Z < z) = 1.75 / 2

= P(Z < z) = 0.875

= P(Z < 1.15) = 0.875

= z  ± 1.15

Using z-score formula,

x = z * +

x = -1.15 * 17 + 250

x = 230.45

Using z-score formula,

x = z * +

x = 1.15 * 17 + 250

x = 269.55

The middle 75% are from 230.45 to 269.55

c) P(x > 275) = 1 - p( x< 275)

=1- p P[(x - ) / < (275 - 250) / 17]

=1- P(z < 1.47)

= 1 - 0.9292

= 0.0708

= 1000 * 0.0708 = 70.8

= 71 cups