Question

widths of solar eclipses the widths in miles of 14 solar eclipses in the united states are shown. find the mean, median, mode, and midrange for the data. 81 70 99 167 118 151 109 184 182 127 257 315 277 298 send data to excel part: 0 / 4 part 1 of 4 (a) find the mean. rounding rule for the mean: round to one more decimal place than the data, as needed. mean: 159 try again

Explanation:

Step1: Sum all the data values

The data values are: 81, 70, 99, 167, 118, 151, 109, 184, 182, 127, 257, 315, 277, 298.
Sum = \(81 + 70 + 99 + 167 + 118 + 151 + 109 + 184 + 182 + 127 + 257 + 315 + 277 + 298\)
Calculate step by step:
\(81+70 = 151\); \(151+99 = 250\); \(250+167 = 417\); \(417+118 = 535\); \(535+151 = 686\); \(686+109 = 795\); \(795+184 = 979\); \(979+182 = 1161\); \(1161+127 = 1288\); \(1288+257 = 1545\); \(1545+315 = 1860\); \(1860+277 = 2137\); \(2137+298 = 2435\)

Step2: Calculate the mean

Mean = \(\frac{\text{Sum of data}}{\text{Number of data points}}\)
Number of data points \(n = 14\)
Mean = \(\frac{2435}{14}\approx173.92857\)
Round to one more decimal place than the data (data has 0 decimal places, so round to 1 decimal place): \(173.9\) (Wait, wait, let's recalculate the sum correctly. Maybe I made a mistake in addition. Let's add again:

81 + 70 = 151

151 + 99 = 250

250 + 167 = 417

417 + 118 = 535

535 + 151 = 686

686 + 109 = 795

795 + 184 = 979

979 + 182 = 1161

1161 + 127 = 1288

1288 + 257 = 1545

1545 + 315 = 1860

1860 + 277 = 2137

2137 + 298 = 2435. Wait, 2435 divided by 14: 14*173 = 2422, 2435 - 2422 = 13, so 173 + 13/14 ≈ 173.92857. But maybe the data was misread? Wait the data is 14 values: let's list all 14:

First row: 81, 70, 99, 167, 118, 151, 109 (7 values)

Second row: 184, 182, 127, 257, 315, 277, 298 (7 values). Total 14.

Wait, maybe I miscalculated the sum. Let's use another approach:

81 + 70 = 151

99 + 167 = 266; 151 + 266 = 417

118 + 151 = 269; 417 + 269 = 686

109 + 184 = 293; 686 + 293 = 979

182 + 127 = 309; 979 + 309 = 1288

257 + 315 = 572; 1288 + 572 = 1860

277 + 298 = 575; 1860 + 575 = 2435. So sum is 2435. Then mean is 2435 /14 = 173.928571... So rounding to one more decimal place than the data (data is whole numbers, so one decimal place) gives 173.9? Wait but maybe I made a mistake in the problem's data? Wait the original problem's data: let's check again.

Wait the user's data: 81,70,99,167,118,151,109,184,182,127,257,315,277,298. Let's count: 14 numbers. So sum is 2435. Then 2435 divided by 14: 14*173 = 2422, 2435-2422=13, so 173 + 13/14 ≈ 173.92857, which is approximately 173.9 when rounded to one decimal place. Wait but the initial wrong answer was 159, so maybe I misread the data. Wait let's check each number:

81, 70, 99, 167, 118, 151, 109, 184, 182, 127, 257, 315, 277, 298. Let's add them in pairs:

(81 + 298) = 379

(70 + 277) = 347; 379 + 347 = 726

(99 + 315) = 414; 726 + 414 = 1140

(167 + 257) = 424; 1140 + 424 = 1564

(118 + 182) = 300; 1564 + 300 = 1864

(151 + 184) = 335; 1864 + 335 = 2199

(109 + 127) = 236; 2199 + 236 = 2435. Yes, sum is 2435. So mean is 2435 /14 ≈ 173.9 (rounded to one decimal place). Wait, maybe the problem had a different data set? Wait no, the user provided the data as 81,70,99,167,118,151,109,184,182,127,257,315,277,298. So sum is 2435, n=14, mean=2435/14≈173.9.

Wait, maybe I made a mistake in the number of data points? Let's count: 81 (1),70(2),99(3),167(4),118(5),151(6),109(7),184(8),182(9),127(10),257(11),315(12),277(13),298(14). Yes, 14 data points. So mean is 2435/14≈173.9.

Answer:

173.9