Question

A technician plans to test a certain type of resin developed in a laboratory to determine the nature of the time it takes before bonding takes place. It is known that the time to bonding is normally distributed with a mean of 4.5 hours and a standard deviation of 1.5 hours. It will be considered an undesirable product if the bonding time is either less than 2 hours or more than 6 hours. a. What is the probability that the bonding time will be less than 2 hours? b. What is the probability that the bonding time will be more than 6 hours? c. How often would the performance be considered undesirable (in a value of probability)? d. Between what two times (equally before the mean and equally after the mean) accounts for the a drying time of 95%?

Answer 1

Solution :

Given that ,

mean = = 4.5 hours

standard deviation = = 1.5 hours

a) P(x < 2) = P[(x - ) / < (2 - 4.5) / 1.5]

= P(z < -1.67)

Using z table,

= 0.0475

b) P(x > 6) = 1 - p( x< 6)

=1- p P[(x - ) / < (6 - 4.5) / 1.5 ]

=1- P(z < 1.00)

Using z table,

= 1 - 0.8413

= 0.1587

c)P(x < 2) + P(x > 6)

= 0.0475 + 0.1587

= 0.2062

d) Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96) = 0.975

= z  ± 1.96

Using z-score formula,

x = z * +

x = -1.96 * 1.5 + 4.5

x = 1.56 hours

Using z-score formula,

x = z * +

x = 1.96 * 1.5 + 4.5

x = 7.44 hours

95% two times is 1.56 hours and 7.44 hours.