The data below indicate the contamination in parts per million in each of 50 samples of drinking water at a specific location. 402
398
416
408
390
412
391
408
406
406
405
420
409
400
406
424
424
420
388
393
413
416
414
411
404
400
407
1650
405
393
409
412
403
401
419
389
396
404
403
404
389
416
402
405
408
407
400
393
403
399

a. What is the first quartile of the data?
b. What is the third quartile of the data?
c. What is the median of the data?
d. What is the mean of the data? Give your answer to three decimal places
e. Values that are greater than Q3 + 1.5 IQR or less than Q1 - 1.5 IQR are typically considered outliers. What value is an outlier in this data?
f. Delete the outlier. What is the median after the outlier is deleted?
g. What is the mean after the outlier is deleted? Give your answer to three decimal places.

Kindly check explanation

Step-by-step explanation:

Given the following :

Sorted data :

388, 389, 389, 390, 391, 393, 393, 393, 396, 398, 399, 400, 400, 400, 401, 402, 402, 403, 403, 403, 404, 404, 404, 405, 405, 405, 406, 406, 406, 407, 407, 408, 408, 408, 409, 409, 411, 412, 412, 413, 414, 416, 416, 416, 419, 420, 420, 424, 424, 1650

n = number of observations = 50

1) first quartile of the data (Q1) : 1/4(n+1)th term

1/4(51)th term= 12.75th term = 400

2)3rd quartile :

3/4(51)th term = 38.25th term = 412

3)median of data (Q2) :

1/2(51)th term = 25.5 th term = 405

4) mean of data

Σx / n = 21501 / 50 = 430.02

5)Outlier : Values that are greater than Q3 + 1.5 IQR or less than Q1 - 1.5 IQR

IQR = Q3 - Q1 = 412 - 400 = 12

> Q3 + 1.5 IQR

> 412 + 1.5(12) = > 430

< Q1 - 1.5 IQR = < 400 - 1.5(12) = 382

Outlier = 1650

Median after outlier is deleted :

1/2(50) th term = 25 th term = 405

Mean after outlier is deleted :

Σx / n = 19851 / 49 = 405.122

answer: add the !

step-by-step explanation: then we will answer

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6