![\bf \displaystyle \cfrac{1}{7}\sum\limits_{i=1}^{7}~a_i\\\\\\ \cfrac{1}{7}\left[ \stackrel{i=1~ a_1}{(2.7)}+\stackrel{i=2~ a_2}{(3.0)}+\stackrel{i=3~ a_3}{(3.2)}+\stackrel{i=4~ a_4}{(3.7)}+\stackrel{i=5~ a_5}{(4.4)}+\stackrel{i=6~ a_6}{(4.9)}+\stackrel{i=7~ a_7}{(5.3)} \right] \\\\\\ \displaystyle \cfrac{1}{7}\sum\limits_{i=1}^{7}~a_i\qquad \approx 3.88571428571428571429\implies \stackrel{\textit{rounded up}}{3.9}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cdisplaystyle%20%5Ccfrac%7B1%7D%7B7%7D%5Csum%5Climits_%7Bi%3D1%7D%5E%7B7%7D~a_i%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B7%7D%5Cleft%5B%20%5Cstackrel%7Bi%3D1~%20a_1%7D%7B%282.7%29%7D%2B%5Cstackrel%7Bi%3D2~%20a_2%7D%7B%283.0%29%7D%2B%5Cstackrel%7Bi%3D3~%20a_3%7D%7B%283.2%29%7D%2B%5Cstackrel%7Bi%3D4~%20a_4%7D%7B%283.7%29%7D%2B%5Cstackrel%7Bi%3D5~%20a_5%7D%7B%284.4%29%7D%2B%5Cstackrel%7Bi%3D6~%20a_6%7D%7B%284.9%29%7D%2B%5Cstackrel%7Bi%3D7~%20a_7%7D%7B%285.3%29%7D%20%5Cright%5D%20%5C%5C%5C%5C%5C%5C%20%5Cdisplaystyle%20%5Ccfrac%7B1%7D%7B7%7D%5Csum%5Climits_%7Bi%3D1%7D%5E%7B7%7D~a_i%5Cqquad%20%5Capprox%203.88571428571428571429%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B3.9%7D%20)
now, le'ts do the summation using 0.5n+2
(0.5(2.7) + 2)+(0.5(3.0) + 2)+(0.5(3.2) + 2)+(0.5(3.7) + 2)+(0.5(4.4) + 2)+(0.5(4.9) + 2)+(0.5(5.3) + 2)
which gives us 27.6. Now, the paper doesn't state, or at least I don't see it, but that's just the sum of the hours, to get the average we simply divide that by 7, the amount of items, and 27.6 ÷ 7 gives us about 3.94.
does it underestimate or overestimate the one found with the summation? well, the summation gave us 3.9 and 3.94 is just 0.04 or 4 hundredths above it.
Answer: 14.555
Step-by-step explanation: so you take $3.85 multiplied by 3 because they bought 3 toy cars plus a $4.56 and take the answer of that and divide by 2 because they split the cost. Hope this helps
Answer:
The lower bound of a 99% C.I for the proportion of defectives = 0.422
Step-by-step explanation:
From the given information:
The point estimate = sample proportion 
= 0.55
At Confidence interval of 99%, the level of significance = 1 - 0.99
= 0.01
Then the margin of error 
E = 0.128156
E ≅ 0.128
At 99% C.I for the population proportion p is: 
= 0.55 - 0.128
= 0.422
Thus, the lower bound of a 99% C.I for the proportion of defectives = 0.422
