Given that Clyde
Clement wants to analyze a shipment of bags of cement. That he knows the
weight of the bags is normally distributed so he can use the standard
normal distribution. That he measured the weight of 600 randomly selected
bags in the shipment. That he calculated the mean and standard
deviation of their weights as 50 lbs and 1.5 lbs respectively.
To
calculate the number of bags in each segment of the distribution we use the normal distribution table.
Case 1: Between -2 and -1 standard deviations.
The mean weight will be -2(1.5) + 50 to -1(1.5) + 50 = -3 + 50 to -1.5 + 50 = 47 lbs to 48.5 lbs
From table, the probability of between -2 and -1 standard deviations = P(-1) - P(-2) = (1 - P(1)) - (1 - P(2)) = P(2) - P(1) = 0.97725 - 0.84134 = 0.13591 ≈ 13.6%
Number of bags out of 600 = 0.13591 x 600 ≈ 82 bags
Case 2: Between -1 and 0 standard deviations.
The mean weight will be -1(1.5) + 50 to 0(1.5) + 50 = -1.5 + 50 to 0 + 50 = 48.5 lbs to 50 lbs
From table, the probability of between -1 and 0 standard deviations = P(0) - P(-1) = P(0) - (1 - P(1)) = P(0) + P(1) - 1 = 0.5 + 0.84134 - 1 = 0.34134 ≈ 34.1%
Number of bags out of 600 = 0.34134 x 600 ≈ 205 bags
Case 3: Between 0 and 1 standard deviations.
The mean weight will be 0(1.5) + 50 to 1(1.5) + 50 = 0 + 50 to 1.5 + 50 = 50 lbs to 51.5 lbs
From table, the probability of between 0 and 1 standard deviations = P(1) - P(0) = 0.84134 - 0.5 = 0.34134 ≈ 34.1%
Number of bags out of 600 = 0.34134 x 600 ≈ 205 bags
Case 4: Between 1 and 2 standard deviations.
The mean weight will be 1(1.5) + 50 to 2(1.5) + 50 = 1.5 + 50 to 3 + 50 = 51.5 lbs to 53 lbs
From table, the probability of between 1 and 2 standard deviations = P(2) - P(1) = 0.97725 - 0.84134 = 0.13591 ≈ 13.6%
Number of bags out of 600 = 0.13591 x 600 ≈ 82 bags
