1.

The weight of an object is normally distributed with a mean of 30 pounds and a standard deviation of 9.8 pounds. A person randomly selected 36 objects and loaded them into a vehicle. What is the probability that the bunch of objects will weigh > 1010 pounds?

Answer»

Response:

Let’s consider a normal distribution N 

This is an example of a central limit theorem. As n increases, the average of x (x-bar let's call) should get closer to the population MEAN (MU) Average of x is DENOTED as x-bar. 

If we consider normal distribution and will try to write below mathematical equation as per normal distribution for this problem: 

X ~ N(30, 9.8/sqrt(36))  Probability of a group of objects WEIGHING > 1010 pounds [Symbol]  Prob(x-bar > 1010/36) = Prob[(x-bar – 30) > (1010/36 - 30)] , which further implies:   Prob(x-bar > 1010/36) = Prob[( (x-bar – 30) / (9.8/sqrt(36) ) ) > ((1010/36 – 30) / (9.8/sqrt(36)) )] =>  Prob(x-bar > 1010/36) = Prob(Z > -1.190)                                      = 1 – 0.1170                                       = 0.883 (as per normal z-score table)  Hence probability is 88.3% or 0.883

Here, the probability that z-score > -1.19 is equal to the blue colour area below of area under the curve. 

Now, the area above -1.19 is same as area below 1.19 as per the distribution diagram below.  


Discussion

No Comment Found

Related InterviewSolutions