Answer:
Option b. None is the correct option.
The Answer is 63%
Step-by-step explanation:
To solve for this question, we would be using the z score formula
The formula for calculating a z-score is given as:
z = (x-μ)/σ,
where
x is the raw score
μ is the population mean
σ is the population standard deviation.
We have boxes X and Y. So we will be combining both boxes
Mean of X = 100 lb
Mean of Y = 5 lb
Total mean = 100 + 5 = 105lb
Standard deviation for X = 1 lb
Standard deviation for Y = 0.5 lb
Remember Variance = Standard deviation ²
Variance for X = 1lb² = 1
Variance for Y = 0.5² = 0.25
Total variance = 1 + 0.25 = 1.25
Total standard deviation = √Total variance
= √1.25
Solving our question, we were asked to find the percent of filled boxes weighing between 104 lb and 106 lb are to be expected. Hence,
For 104lb
z = (x-μ)/σ,
z = 104 - 105 / √25
z = -0.89443
Using z score table ,
P( x = z)
P ( x = 104) = P( z = -0.89443) = 0.18555
For 1061b
z = (x-μ)/σ,
z = 106 - 105 / √25
z = 0.89443
Using z score table ,
P( x = z)
P ( x = 106) = P( z = 0.89443) = 0.81445
P(104 ≤ Z ≤ 106) = 0.81445 - 0.18555
= 0.6289
Converting to percentage, we have :
0.6289 × 100 = 62.89%
Approximately = 63 %
Therefore, the percent of filled boxes weighing between 104 lb and 106 lb that are to be expected is 63%
Since there is no 63% in the option, the correct answer is Option b. None.
