2. Manufacturer A produces hammers that are normally distributed with a mean weight of 4.6 lb and a standard deviation of 0.8 lb. Manufacturer B produces hammers that are normally distributed with a mean weight of 6.3 lb and a standard deviation of 1.4 lb.
(a) What percentage of Manufacturer A’s hammers will weigh less than 5 lb?
(b) What percentage of Manufacturer B’s hammers will weigh less than 5 lb?
(c) Which manufacturer is more likely to produce a hammer weighing exactly 5.2 lb? Explain.
http://davidmlane.com/hyperstat/z_table.html
To answer these questions, we need to use the concept of the normal distribution and z-scores.
(a) To find the percentage of Manufacturer A's hammers that will weigh less than 5 lb, we can calculate the z-score first. The z-score measures the number of standard deviations an observation is from the mean. The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
x = the value we want to find the percentage for (5 lb in this case)
μ = the mean weight of Manufacturer A's hammers (4.6 lb)
σ = the standard deviation of Manufacturer A's hammers (0.8 lb)
Plugging in the values, we get:
z = (5 - 4.6) / 0.8 = 0.5
Next, we need to use a z-table or a statistical calculator to find the percentage of Manufacturer A's hammers that will weigh less than 5 lb. From the z-table, we find that the percentage corresponding to a z-score of 0.5 is approximately 0.6915 or 69.15%.
Therefore, approximately 69.15% of Manufacturer A's hammers will weigh less than 5 lb.
(b) Following the same steps as in part (a), we need to calculate the z-score for Manufacturer B's hammers using the values provided:
z = (5 - 6.3) / 1.4 = -0.9286
Using the z-table, we find that the percentage corresponding to a z-score of -0.9286 is approximately 0.1772 or 17.72%.
Thus, approximately 17.72% of Manufacturer B's hammers will weigh less than 5 lb.
(c) To determine which manufacturer is more likely to produce a hammer weighing exactly 5.2 lb, we compare the z-scores for both manufacturers.
For Manufacturer A:
z_A = (5.2 - 4.6) / 0.8 = 0.75
For Manufacturer B:
z_B = (5.2 - 6.3) / 1.4 = -0.7857
From the z-scores, we can see that Manufacturer A's z-score is higher (closer to the mean). This implies that the weight of 5.2 lb is relatively closer to the mean weight of Manufacturer A's hammers compared to Manufacturer B's hammers. Therefore, Manufacturer A is more likely to produce a hammer weighing exactly 5.2 lb.