Download 1) A food processor packages orange juice in small jars. The

1) A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally
distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounces. Find the proportion of all jars
packaged by this process that have weights that fall above 10.95 ounces.
z = (x – m)/s = (10.95 – 10.5)/0.3 = 1.5
P(x > 10.95) = P(z > 1.5) = 0.067
Proportion of the jars weighing more than 10.95 ounces = 0.067
2) A catalog company that receives the majority of its orders by telephone conducted a study to determine how long
customer were willing to wait on hold before ordering a product. The length of time was found to be a random variable
best approximated by an exponential distribution with a mean equal to 3 minutes. Find the waiting time at which only
10% of the customer will continue to hold.
P(X > x) = e^(-x) = e^(-x/3) = 0.10
On solving, we get x = 6.91 minutes
3) Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the
probability that a randomly chosen arrival would be more than 1 hour?
P(x > 1) = e^(-1) = 0.368
4) The time needed to complete a final exam in a particular college course is normally distributed with a mean of 80
minutes and a standard deviation of 10 minutes. If there is 90 minutes allotted for the exam, what percentage of
students do you expect will be unable to complete the exam?
z = (x – m)/s = (90 – 80)/10 = 1
P(x > 90) = P(z > 1) = 0.1587
15.87% of the students will be unable to complete the exam
5) If Z has a standardized normal distribution, what is the probability that Z is more than
-0.98 is?
P(z > -0.98) = 0.8365
6)The amount of time necessary for assembly line workers to complete a product is a normal random variable with a
mean of 15 minutes and a standard deviation of 2 minutes. 17% of the products would be assembled within???
minutes
Area under the standard normal curve is 17% that is 0.17 to the left of z = -0.9542
x = m + z * s = 15 + (-0.9542)(2) = 13.09
17% of the products would be assembled within 13.09 minutes
7)Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the
probability that a randomly chosen arrival would be less than 20 minutes?
P(x < 20) = 1 – e^(-20 * 1/60) = 0.283
8) Let X represent the amount of time it takes a student to park in the library parking lot at the university. If the
distribution of parking times can be modeled using an exponential distribution with a mean of 4 minutes, find the
probability that it will take a randomly selected student more than 10 minutes to park in the library lot.
P( x > 10) = e^ (- 10/4) = 0.082
9) A catalog company that receives the majority of its orders by telephone conducted a study to determine how long
customers were willing to wait on hold before ordering a product. The length of time was found to be a random
variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What proportion of
callers is put on hold longer than 2.8 minutes?
P(x > 2.8) = e^ (- 2.8/2.8) = 0.3679.