Question

the weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5 oz. and a standard deviation of 1 oz. using the empirical rule, determine what interval would represent weights of the middle 95% of all oranges from this orchard.

Explanation:

Step1: Recall the empirical rule

The empirical rule for a normal distribution states that approximately 95% of the data lies within \( \mu\pm2\sigma\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation.

Step2: Identify the mean and standard deviation

Given \(\mu = 7.5\) oz (mean weight of oranges) and \(\sigma=1\) oz (standard deviation of orange weights).

Step3: Calculate the lower and upper bounds of the interval

For the lower bound: \(L=\mu - 2\sigma\). Substitute \(\mu = 7.5\) and \(\sigma = 1\), we get \(L=7.5-2\times1=7.5 - 2=5.5\).
For the upper bound: \(U=\mu + 2\sigma\). Substitute \(\mu = 7.5\) and \(\sigma = 1\), we get \(U=7.5 + 2\times1=7.5+2 = 9.5\).

Answer:

The interval representing the weights of the middle 95% of all oranges is \((5.5,9.5)\) oz.