The weight of lolly pops in a box is normally distributed with a mean of 5kg and a standard deviation of 0.1kg. What is the probability that a randomly selected box of lolly pops weighs between 5.03 and 5.16 kg?
(5.03-5.0)/0.1 = 0.3 S.D.
(5.16-5.0)/0.1 = 1.6 S.D.
Look up the z tables to find what probability (of a normal distribution) is between 0.3 and 1.6 standard deviations.
I get 0.4452-0.1179=0.3273
To find the probability that a randomly selected box of lolly pops weighs between 5.03 and 5.16 kg, we need to calculate the area under the normal distribution curve between these two values.
First, let's standardize the values using the formula:
Z = (X - μ) / σ
Where:
X = the value we want to standardize (in this case, 5.03 kg and 5.16 kg)
μ = the mean of the distribution (5 kg)
σ = the standard deviation of the distribution (0.1 kg)
For 5.03 kg:
Z1 = (5.03 - 5) / 0.1 = 0.3
For 5.16 kg:
Z2 = (5.16 - 5) / 0.1 = 1.6
Next, we need to find the probability associated with these standardized values.
Using a standard normal distribution table or a statistical software, we can determine the probability associated with each Z-value.
Referencing the standard normal distribution table, the probability associated with Z = 0.3 is approximately 0.6179, and the probability associated with Z = 1.6 is approximately 0.9452.
Now, to find the probability between these two Z-values, we subtract the probability associated with the lower Z-value from the probability associated with the higher Z-value:
P(0.3 < Z < 1.6) = P(Z < 1.6) - P(Z < 0.3)
P(0.3 < Z < 1.6) = 0.9452 - 0.6179
P(0.3 < Z < 1.6) ≈ 0.3273
Therefore, the probability that a randomly selected box of lolly pops weighs between 5.03 and 5.16 kg is approximately 0.3273 or 32.73%.