Question

**Potatoes:** Suppose the weights of Farmer Carl's
potatoes are normally distributed with a mean of 8 ounces and a
standard deviation of 0.9 ounces.

(a) Carl only wants to sell the best potatoes to his friends and
neighbors at the farmer's market. According to weight, this means
he wants to sell only those potatoes that are among the heaviest
25%. What is the minimum weight required to be brought to the
farmer's market? **Round your answer to 2 decimal
places.**

ounces

(b) He wants to use the lightest potatoes as ammunition for his
potato launcher but can only spare about 5% of his crop for such
frivolities. What is the weight limit for potatoes to be considered
for ammunition? **Round your answer to 2 decimal
places.**

ounces

(c) Determine the weights that delineate the middle 95% of Carl's
potatoes from the others. **Round your answers to 2 decimal
places.**

from __ to __ ounces

Answer #1

**Answer:**

Given,

Mean = 8

Standard deviation = 0.9

a)

To give the minimum weight required to be brought to the farmer's market

P(z > 0.25) = 0.6745

since from standard normal table

(x - 8)/0.9 = 0.6745

x - 8 = 0.6745*0.9

x = 8 + 0.6745*0.9

x = 8.61 ounces

b)

To give the weight limit for potatoes to be considered for ammunition

P(Z < z) = 5%

P(z < - 1.65) = 0.05

since from standard normal table

z = - 1.65

consider,

x = z*s + u

substitute values

x = -1.65*0.9 + 8

x = -1.485 + 8

x = 6.52 ounces

c)

To give middle 95% , z value is given below

At 95% confidence interval, z value is 1.96

Limits = (xbar - u)/s = +/- 1.96

substitute values

Lower limit ,

(XL - 8)/0.9 = - 1.96

XL - 8 = -1.96*0.9

XL = 8 - 1.764

XL = 6.236

Upper limit ,

(XU - 8)/0.9 = 1.96

XU - 8 = 1.96*0.9

XU = 8 + 1.96*0.9

XU = 9.764

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