Your local grocery store sells 2 lb. bags of Jasmine rice. However, the 2 lb. ba

Question

Your local grocery store sells 2 lb. bags of Jasmine rice. However, the 2 lb. bags don’t weigh exactly 2 lbs. If we let X be the weight of a randomly selected 2 lb. bag of Jasmine rice, historical data indicates that X sub i~N(mean = 2.41, standard deviation =0.21). The local warehouse store sells 10 lb. bags of Jasmine rice, which also do not weight exactly 10 lbs. If Y is the weight of a randomly selected 10 lb. bag, historical data indicates that Y~N(mean= 10.32, standard dev = 0.30). Suppose we randomly select five 2 lb. bags of Jasmine rice from the local store and one 10 lb. bag of Jasmine rice from the warehouse store. Assume the weights of all bags are independent.

a) Let L = the total weight of the five randomly selected 2 lb. bags of Jasmine rice from the local store. What is the distribution of L? Be sure to name the distribution and parameter values, showing any necessary calculations.

b) Define a new random variable W = L – Y. What is the distribution of W? Be sure to name the distribution and parameter values, showing all calculations.

c) What is the probability that the sum of the weights of the five 2 lb. bags of Jasmine rice exceeds the weight of one 10 lb. bag?

Explanation / Answer

Let X be the random variable that the weight of a randomly selected 2 lb. bag of Jasmine rice.

Given that,

X ~ N(mean = 2.41, sd = 0.21)

Let Y be the random variable that the weight of a randomly selected 10 lb. bag.

Y ~ N(mean = 10.32, sd=0.30)

Suppose we randomly select five 2 lb. bags of Jasmine rice from the local store and one 10 lb. bag of Jasmine rice from the warehouse store. Assume the weights of all bags are independent.

a) Let L = the total weight of the five randomly selected 2 lb. bags of Jasmine rice from the local store. What is the distribution of L? Be sure to name the distribution and parameter values, showing any necessary calculations.

We know that sum of normal random variables is also normal.

L ~ N(mean = 5*2.41, sd = 5*0.21)

L ~ N(mean = 12.05, sd = 1.05)

We know that the difference of L and Y also follows normal distribution with

mean = 12.05-10.32 = 1.73

sd = 1.05 + 0.30 = 1.35

Therefore L-Y ~ N(mean = 1.73, 1.35)

c) What is the probability that the sum of the weights of the five 2 lb. bags of Jasmine rice exceeds the weight of one 10 lb. bag?

Here we have to find P(L > Y).

P(L > Y) = P(L- Y > 0) = P(W >0)

convert W = 0 into z-score

z-score is defined as,

z = (w - mean) / sd

z = ( 0 - 1.73) / 1.35 = -1.28

Now we have to find P( Z > -1.28)

This probability we can find by using EXCEL.

syntax :

=1 - NORMSDIST(z)

where z is the z-score.

P(Z > -1.28) = 0.9000

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