the mass of cherries grown on a farm in the okanogan valley are normally distributed with a mean of 7.0…

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.