the masses of lettuces sold at a market stall are normally distributed with the mean 600g and standard deviation 20g.
1) find the mass exceeded by 10% of the lettuces.
2) 5% of the lettuces have a mass less than M g. find M
1) Well, let's start by finding the z-score corresponding to the 10th percentile (10% of the lettuces will have a mass less than this value). The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x = the value we want to find
μ = mean
σ = standard deviation
The z-score for the 10th percentile is -1.28 (you can reference a z-table or use a calculator). Now we can solve for x:
-1.28 = (x - 600) / 20
Rearranging the equation:
-25.6 = x - 600
Solving for x:
x = 600 - 25.6
x ≈ 574.4 g
So, 10% of the lettuces sold at the market stall weigh less than approximately 574.4 grams.
2) Similarly, we are looking for the mass (M) below which 5% of the lettuces fall. We need to find the z-score corresponding to the 5th percentile:
z = (M - μ) / σ
The z-score for the 5th percentile is -1.645. Now we can solve for M:
-1.645 = (M - 600) / 20
-32.9 = M - 600
M = 600 - 32.9
M ≈ 567.1 g
So, approximately 5% of the lettuces sold at the market stall weigh less than 567.1 grams.
To solve these problems, we can use the concept of the standard normal distribution, also known as the Z-score.
1) To find the mass exceeded by 10% of the lettuces, we need to find the Z-score corresponding to that percentile and then convert it back to the original mass value.
Step 1: Calculate the Z-score for the 10th percentile.
Z = invNorm(0.10) [Using a Z-table or statistical software]
Z ≈ -1.28
Step 2: Convert the Z-score back to the original mass value.
Z = (X - Mean) / Standard Deviation
-1.28 = (X - 600) / 20
Solving for X:
-1.28 * 20 = X - 600
-25.6 = X - 600
X ≈ 574.4
So, the mass exceeded by 10% of the lettuces is approximately 574.4g.
2) To find the mass corresponding to the 5th percentile, we can follow a similar approach.
Step 1: Calculate the Z-score for the 5th percentile.
Z = invNorm(0.05)
Z ≈ -1.645
Step 2: Convert the Z-score back to the original mass value.
-1.645 = (X - 600) / 20
Solving for X:
-1.645 * 20 = X - 600
-32.9 = X - 600
X ≈ 567.1
Therefore, 5% of the lettuces have a mass less than approximately 567.1g.
To find the answer to these questions, we can use the properties of the normal distribution and the z-score formula. The z-score measures the number of standard deviations an observation is from the mean.
1) To find the mass exceeded by 10% of the lettuces, we need to find the z-score that corresponds to a cumulative probability of 90% (100% - 10%). We'll use a z-table or a statistical calculator to look up the z-score associated with a cumulative probability of 0.9.
The z-score can be calculated using the formula:
z = (x - mean) / standard deviation
Since we need to find the mass exceeded, we want the z-score associated with the upper tail of the distribution. Therefore, we'll use the formula:
z = invNorm(0.9) * standard deviation + mean
Substituting the values given in the question, we have:
z = invNorm(0.9) * 20 + 600
Using a z-table or a statistical calculator, we find that invNorm(0.9) ≈ 1.28.
z = 1.28 * 20 + 600
z ≈ 625.6
Therefore, approximately 10% of the lettuces have a mass greater than 625.6g.
2) To find the mass M such that 5% of the lettuces have a mass less than M, we need to find the z-score that corresponds to a cumulative probability of 5%. Again, we'll use a z-table or a statistical calculator to find the z-score associated with a cumulative probability of 0.05.
The z-score can be calculated using the formula:
z = (x - mean) / standard deviation
Since we need to find the mass less than M, we want the z-score associated with the lower tail of the distribution. Therefore, we'll use the formula:
z = invNorm(0.05) * standard deviation + mean
Substituting the values given in the question, we have:
z = invNorm(0.05) * 20 + 600
Using a z-table or a statistical calculator, we find that invNorm(0.05) ≈ -1.64.
z = -1.64 * 20 + 600
z ≈ 568.8
Therefore, approximately 5% of the lettuces have a mass less than 568.8g.