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2. For the data of the previous problem, determine the range within which observations should fall: a) 90% of the time. and b) 95% of the time. Also, list the percentages of values that actually fall within these ranges. 1. Distance AB is observed repeatedly using the same equipment and procedures, yielding the following results in meters: 26.319, 26.323, 26.310, 26.320, 26.318, 26.313, 26.316, 26.318, 26.328, and 26.325. Find: a) the line's most probable length, b) the standard deviation, and c) the standard deviation of the mean.

Answer 1

Ans) Total number of observations(n) = 10

Sum of observations = 26.319 + 26.323......26.235

=> Sum =26 + 3.19

Mean (x') = 26 + 3.19 / n

= 26 + 3.19/10

= 26.319

Observed Length	V = x' - x	V2
26.319	0	0
26.323	-0.004	0.000016
26.310	0.009	0.000081
26.320	-0.001	0.000001
26.318	0.001	0.000001
26.313	0.006	0.000036
26.316	0.003	0.000009
26.318	0.001	0.000001
26.328	-0.009	0.000081
26.325	-0.006	0.000036
	$\sum_{}^{}$V = 0	$\sum_{}^{}$V2 = 0.000262

Standard deviation ($\sigma$) = [V² / (n-1) ]^0.5

= 0.0054

Now, lower confidence limit =x' - t $\sigma$ /n^0.5

Ans a) From statistical table, for 90% confidence , t= 1.833

Lower limit = x' - 1.833 x 0.0054 / 10^0.5

= 0.00313

Lower value = x' - 0.00313

= 26.319 - 0.00313

= 26.3158

Higher limit = x' + 0.00313

= 26.3221

So, required range is 26.3158 < x < 26.3221

  Ans b)  From statistical table, for 90% confidence , t= 2.26

Lower value = x' - 2.26(0.0054)/10^0.5

= 26.319 - 0.00385

= 26.3151

Higher value = x' + 0.00385

= 26.319 + 0.00385

=26.3228

So required range is, 26.3151 < x < 26.3228