Question

the mass of cherries grown on a farm in the okanogan valley are normally distributed with a mean of 7.0 grams and a standard deviation of 1.1 grams.
a) determine the interval of cherry weights symmetrical about the mean, in which you would expect the mass of 90% of the cherries to be found.
b) the smallest 10% of the cherries will not be sold. what is the minimum selling mass?

Explanation:

Step1: Recall the empirical rule for normal distribution

For a normal distribution, approximately 95% of the data lies within \( \mu\pm2\sigma \), and approximately 99.7% lies within \( \mu\pm3\sigma \). For a symmetric interval about the mean for 90% of the data, we use the z - score. The z - score \( z\) for a two - tailed 90% confidence interval is \( z = 1.645\) (from standard normal distribution tables). The formula for the interval is \( \mu\pm z\sigma\). Given \( \mu = 7.0\) grams and \( \sigma=1.1\) grams.

Step2: Calculate the interval for part (a)

The lower bound of the interval is \( \mu - z\sigma\) and the upper bound is \( \mu + z\sigma\).
Substitute the values: \( 7.0-1.645\times1.1\) and \( 7.0 + 1.645\times1.1\).
First, calculate \( 1.645\times1.1=1.8095\).
Then, \( 7.0-1.8095 = 5.1905\) and \( 7.0+1.8095=8.8095\). The interval is \((5.19,8.81)\) (rounded to two decimal places).

Step3: Use the z - score for part (b)

We want to find the value \( x\) such that \( P(X\lt x)=0.10\). Looking up the z - score in the standard normal distribution table for a left - tailed probability of 0.10, the z - score \( z=- 1.28\).
Using the formula \( z=\frac{x-\mu}{\sigma}\), we can solve for \( x\). Rearranging the formula gives \( x=\mu+z\sigma\).
Substitute \( \mu = 7.0\), \( z=-1.28\), and \( \sigma = 1.1\).
\( x=7.0+(-1.28)\times1.1=7.0 - 1.408=5.592\approx5.59\) grams.

Answer:

a) The interval is \((5.19,8.81)\) grams.
b) The minimum selling mass is \(5.59\) grams.