a random variable having a normal distribution with σ = 0.05 ounce, find the mean such that only 1% of the bags contain less than 8 ounces of potato chips.
To find the mean for which only 1% of the bags contain less than 8 ounces of potato chips, we can use the inverse normal distribution.
The inverse normal distribution is used to find the value of a variable given the probability. In this case, we want to find the value of the variable (which is the mean) for which only 1% of the bags contain less than 8 ounces of potato chips.
Here's how we can calculate it:
1. Determine the z-score corresponding to the desired probability of 1%. The z-score represents the number of standard deviations away from the mean. To find this, use a standard normal distribution table or an online calculator. In this case, we're looking for the z-score that corresponds to a cumulative probability of 1% in the left tail of the distribution.
The z-score for a 1% left tail is approximately -2.33.
2. Use the formula for the z-score:
z = (x - μ) / σ
where z is the z-score, x is the observed value, μ is the mean, and σ is the standard deviation.
In this case, we have:
-2.33 = (8 - μ) / 0.05
3. Solve for the mean (μ):
-2.33 × 0.05 = 8 - μ
-0.1165 = 8 - μ
μ = 8 + 0.1165
μ ≈ 8.12
Therefore, the mean such that only 1% of the bags contain less than 8 ounces of potato chips is approximately 8.12 ounces.