1) Record the following for your histograms for length and width. (For each numb

Question

1) Record the following for your histograms for length and width. (For each number of measurements, enter an integer.)

a) number of measurements within 1 standard deviation from the mean

N1l=   15

N1w=   21

b) number of measurements within 2 standard deviations from the mean

N1l= 29

N1w= 29

Calculate the percentage of the length and width measurements that fall in the range of one standard deviation from the mean value. (Enter your answer to at least three significant figures.)

P1l=   ___________

P1w= ____________

Calculate the percentage of the length and width measurements that fall in the range of two standard deviations from the mean value. (Enter your answer to at least three significant figures.)

P2l= ____________

P2w=____________

Do your measurements appear normally distributed? (Assume the measurements are normally distributed if both of the following are true: that between 60% and 75% of the measurements are within one standard deviation of the mean, and that between 90% and 100% of the measurements are within two standard deviations of the mean.)

EXPLAIN:

Explanation / Answer

For Length, total number of measurements N_{LT} = N1l + N2l = 15 + 29 = 44.  

For weight, total number of measurements N_{WT} = N1w + N2w = 21 + 29 = 50.

Therefore, P1l = (N1l / N_{LT}) * 100 = (15/44) * 100 = 34.091 %

And , P1w = (N1w / N_{WT}) * 100 = (21/50) * 100 = 42.000 %

Similarly, P2l = (N2l / N_{LT}) * 100 = (29/44) * 100 = 65.909 %

And, P2w = (N2w / N_{WT}) * 100 = (29/50) * 100 = 58.000 % ;

Since P1l < 60% , P1w< 75% , P2l <90% and P2w <100% , none of the conditions are satisfied. Hence, the measurements are not normally distributed.

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