Jobs and productivity! how do banks rate? one way to answer this question is to examine annual profits per employee. forbes top companies, edited by j. t. davis (john wiley & sons), gave the following data about annual profits per employee (in units of one thousand dollars per employee) for representative companies in financial services. companies such as wells fargo, first bank system, key banks, norwest banks, and so on were included.

42.9 43.8 48.2 60.6 54.9 55.1 52.9 54.9 42.5 33.0 33.6 36.9 27.0 47.1 33.8 28.1 28.5 29.1 36.5 36.1 26.9 27.8 28.8 29.3 31.5 31.7 31.1 38.0 32.0 31.7 32.9 23.1 54.9 43.8 36.9 31.9 25.5 23.2 29.8 22.3 26.5 26.7

(a) verify that for the preceding data ¯x = 36.0 and s = 10.2 approximately.

(b) let us say that the preceding data are representative of the entire sector of (successful) financial services corporations. find a 75% confidence interval for µ, the average annual profit per employee for all successful banks.

(c) let us say that you are the manager of a local bank with a large number of employees. suppose the annual profits per employee are less than 30 thousand dollars per employee. do you think that this might be somewhat low compared with other successful financial institutions? explain by referring to the confidence interval you computed in (b).

(d) suppose the annual profits are more than 40 thousand dollars per employee. as manager of the bank, would you feel somewhat better? explain by referring to the confidence interval you computed in (b).

(e) repeat (b), (c), and (d) for a 90% confidence level.

See explanation below

Step-by-step explanation:

a)  

Effectively,  

s = 10.2376 ≅ 10.24

b)

Since the sample size is 42, we must find a values c,d such that the area inside the interval [v, w] of the Normal curve with mean 36 and standard deviation is 75% of 1 = 0.75

(See picture 1 attached)

We can easily work out that value with the help of a spreadsheet and we find

[v,w] = [34.182, 37.818]  

c)

Less than 30 thousand dollars per employee means they are among the 25/2 = 12.5% of worst paid employees. So, this could be considered low.

d)

If the annual profits are more than 40 thousand dollars they are among the 12.5% best paid, so they much better off.

e)

If we want now a 90% confidence interval, we repeat the above steps, but now the area inside the interval [v, w] must be 90% or 0.9

Again with the help of the computer

[v, w] = [33.401, 38.599] (See picture 2)

If the employees earn less than 30,000 they are among the worst paid 5% of the interval and if they won more than 40,000 they would be among the highest 5% of the interval  

i don't see any

step-by-step explanation:

the given growth  function for the length of the fish (cm) is

f(m) = 5(1.07)m = 5.35m

where 

m = months

at the end of the study, the length is 9.19 cm. the  time for conducting the study is determined as

5.35m = 9.19

m = 1.7178 months

part a.

the study was performed over a time period  of about 1.72 months. therefore  the domain for  plotting the function is 0 < = m < = 1.72.

note:

if growth  continues at this rate for up to m months, then the domain would be 0< =m< =m.

part b.

the growth function is a linear function.

when m=0, f=0.

therefore the initial growth is zero, which represents the y-intercept.

part c.

assume that the growth function is valid for 0< =m< =9.

the average rate of change of the function is the slope of the linear function, which is 5.35 cm/month.