Question

A few years in the future, tax cuts have created a booming economy. You run a small tree planting business out of Terrace and have become concerned that you will have trouble recruiting students to plant this summer. Tree planters are paid per tree planted. You review data for the past few seasons and discovered that the average tree planter earns $95 per day with a standard deviation of $20. The distribution of payments is normally distributed.

(a)Employment forecasts indicate that most students will not accept a job that pays less than $70 per day. What percentage of your planters make less than this minimum?

(b)You decide to increase the average wage you pay to your planters by paying them a fixed daily allowance in addition to the money they earn planting. How large of an allowance should you pay so that 95% of planters make $70.00 or more per day?

(c)Another company has decided to pay more for each tree planted. They expect a new mean of $105 and a standard deviation of $25 (the distribution is normally distributed). A random sample of 20 planters showed an average wage of $110. If the expected mean and standard deviation are correct, what percentage of sample means of size 20 would be less than $110?

Answer 1

(a)

X ~ N(95, 20)

Percentage of your planters make less than this minimum

= P(X < 70) = P[Z < (70 - 95)/20] = P[Z < -1.25] = 0.1056 = 10.56%

(b)

Let k be the fixed daily allowance in addition to the money .

Then planters earn Y = X + k

E(Y) = E(X + k) = E(X) + k = 95 + k

Var(Y) = Var(X + k) = Var(X) + 0 = 20^2

SD(Y) = 20

Thus, Y ~ N(95 + k, 20)20

Z score for 95% is 1.645.

95% of planters make $70.00 or more per day, so,

70 = (95 + k) + 1.645 * 20

k = (70 - 95) + 1.645 * 20 = $7.9

(c)

Standard error of mean = 25 / = 5.59

~ N(105, 5.59)

Percentage of sample means of size 20 would be less than $110

= P( < 110)

= P[Z < (110 - 105)/5.59]

= P[Z < 0.8945]

= 0.8145 = 81.45%