Instant dinner comes in packages with weights that are normally distributed, with a standard deviation of 0.3 oz. If 2.3% of the dinners weigh more than 13.5 oz, what is the mean weight?
To find the mean weight of the instant dinners, we can use the concept of the standard normal distribution.
The standard normal distribution has a mean of 0 and a standard deviation of 1. By standardizing the data, we can convert it to the standard normal distribution.
Let's assume the mean weight of the instant dinners is μ oz.
We know that 2.3% of the dinners weigh more than 13.5 oz, which means this data falls in the upper tail of the distribution. We need to find the z-score corresponding to the 2.3% percentile.
To find the z-score, we refer to the standard normal distribution table or use a statistical calculator. The 2.3% percentile corresponds to a z-score of approximately 1.81.
The formula for standardizing the data is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value in question (13.5 oz, in this case)
- μ is the mean
- σ is the standard deviation
Substituting the values into the equation:
1.81 = (13.5 - μ) / 0.3
Now, we can solve for μ:
1.81 * 0.3 = 13.5 - μ
0.543 = 13.5 - μ
μ = 13.5 - 0.543
μ ≈ 12.957
Therefore, the mean weight of the instant dinners is approximately 12.957 oz.