Instant dinner comes in packages with weights that are normally distributed with a standard deviation of 0.3 oz. If 2.3 percent of the dinners weigh more than 13.5 oz, what is the mean weight?
12.9 oz, which is two standard deviations below 13.5 oz
See http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html for a tool to do problems like this
2.3% = .023
so that is the probability that it will be over 13.5 oz, or .977 is the prob that it will be below 13.5
normal distribution charts show the probability below a given z-score.
z-score = (real score - mean)/standard deviation
I used this chart
http://www.math.unb.ca/~knight/utility/NormTble.htm
I found .977 at a z-score of 2.00
so 2.00 = (13.5-m)/.3
m = 12.9
Using this data in
http://davidmlane.com/hyperstat/z_table.html
and clicking on "above" at 13.5 gave me .02275, close enough for .023
Well, it seems like these instant dinner packages are quite weighty! Let's calculate the mean weight, shall we?
Since we know that the standard deviation is 0.3 oz, we can find the z-score corresponding to the 2.3 percent of dinners weighing more than 13.5 oz. Using a standard normal distribution table (or a calculator), we can find that the z-score for this probability is approximately 1.811.
Now, let's use the z-score formula: z = (x - μ) / σ, where z is the z-score, x is the value (13.5 oz in this case), μ is the mean weight, and σ is the standard deviation. Plugging in the known values, we have:
1.811 = (13.5 - μ) / 0.3
Now, we can solve for the mean weight (μ):
1.811 * 0.3 = 13.5 - μ
0.5433 = 13.5 - μ
μ = 13.5 - 0.5433
μ ≈ 12.9567 oz
Therefore, the mean weight of these instant dinners is approximately 12.96 oz. Just be careful not to accidentally launch them into space while opening the package!
To find the mean weight of the instant dinners, we can use the standard normal distribution table. First, we need to find the z-score associated with the given percentage.
The z-score formula is given by:
z = (x - μ) / σ
Where:
z = z-score
x = observed value (13.5 oz in this case)
μ = mean weight
σ = standard deviation (0.3 oz in this case)
Rearranging the formula to solve for μ:
μ = x - (z * σ)
Now, let's find the z-score using the cumulative probability function.
P(Z > z) = 2.3% = 0.023 (since we need to find the percentage above this value)
From the standard normal distribution table, we find that the z-score for a cumulative probability of 0.023 is approximately -1.84.
Substituting the values into the formula:
μ = 13.5 - (-1.84 * 0.3)
= 13.5 + 0.552
= 14.052 oz
Therefore, the mean weight of the instant dinners is approximately 14.052 oz.
To determine the mean weight of the instant dinners, we can use the concept of standard deviation and the Z-score.
1. Let's define the variables:
- X: Random variable representing the weight of the instant dinners
- μ: Mean weight (unknown)
- σ: Standard deviation = 0.3 oz
- Z: Z-score
2. We are given that 2.3 percent of the dinners weigh more than 13.5 oz. To find the Z-score corresponding to this value, we will use a standard normal distribution table or a calculator.
Using a standard normal distribution table, we find that the Z-score corresponding to 2.3 percent (or 0.023) is approximately 1.88.
3. The Z-score formula relates the observed value (X) to the mean (μ) and the standard deviation (σ):
Z = (X - μ) / σ
Rearranging the formula, we have:
X = Z * σ + μ
Substituting the known values:
X = 1.88 * 0.3 + μ
4. Now, we can solve for the mean weight (μ). Since 13.5 oz is the observed value (X), we can plug it into the formula:
13.5 = 1.88 * 0.3 + μ
Simplifying the equation:
13.5 = 0.564 + μ
Moving μ to the left side of the equation:
μ = 13.5 - 0.564
μ ≈ 12.936
Therefore, the mean weight of the instant dinners is approximately 12.936 oz.