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: Find z - scores for 90% symmetric interval

For a 90% symmetric interval about the mean in a normal distribution, the area in the two tails is \(1 - 0.90=0.10\). The area in each tail is \(\frac{0.10}{2}=0.05\). Using the standard normal distribution table (z - table), the z - score \(z_{1}\) corresponding to an area of \(0.05\) in the left - tail is \(z=- 1.645\) and the z - score \(z_{2}\) corresponding to an area of \(0.95\) (since \(1 - 0.05 = 0.95\)) in the right - tail is \(z = 1.645\).
The formula for a value \(x\) in a normal distribution is \(x=\mu+z\sigma\), where \(\mu = 7.0\) grams (mean) and \(\sigma=1.1\) grams (standard deviation).

Step2: Calculate the lower and upper bounds of the 90% interval

For the lower bound (\(z=-1.645\)):
\(x_{1}=\mu+z\sigma=7.0+(-1.645)\times1.1=7.0 - 1.8095 = 5.1905\)
For the upper bound (\(z = 1.645\)):
\(x_{2}=\mu+z\sigma=7.0 + 1.645\times1.1=7.0+1.8095 = 8.8095\)
The interval is \((5.1905,8.8095)\)

Step3: Find the z - score for the 10% left - tail

For the 10% left - tail (the smallest 10% of the cherries), the area to the left of the z - score is \(0.10\). Using the z - table, the z - score \(z\) corresponding to an area of \(0.10\) is approximately \(z=-1.28\)
Using the formula \(x=\mu+z\sigma\) with \(\mu = 7.0\) and \(\sigma = 1.1\)
\(x=7.0+(-1.28)\times1.1=7.0-1.408 = 5.592\)

Answer:

a) The interval of cherry weights is \((5.19,8.81)\) grams (rounded to two decimal places)
b) The minimum selling mass is \(5.59\) grams (rounded to two decimal places)