Question 4 of 8 Using Chebysher\'s theorem, solve these problems for a distribut

Question

Question 4 of 8 Using Chebysher's theorem, solve these problems for a distribution with a mean of 69 and a standard deviation of &. Round & and final anwers to one decimal place. Part 1 At least of the valees will fall between 37 and 101. Part 2 Ar least of the valaes will fall between 42 and 96 Question 5 of 8 Americans spend an average of 4 hours per day online. If the standard deviation is 31minutes,ind the range which at least 93.75% of the data will be. Use Cbebysbrr's theorem. Round youuk the pearest whole mumber. [ ]and At least 93.5% ofthe data Till lie between m-es. ALEKS Eye London Question 6 of 8 The average sale price of ew one family houses in the Uaited States for a recest year was $249 3300 Find the range ofraluein which at least 93."5% ofthe sale price. w., be ifth, saadard dniation s $47,400 Round yourk to the nearest wbole sumber The range of values is beew een Question 7 of 8 The national average for mathematics on a standardized test in 2011 was S08 Sappose that the distribution of scores was approximately bell-shaped and that the standard deviacion wa approsimately 40. Round your aniwers to one decimal place as seeded Part Within what boundaries would you espect 6596 of the scores to fall? €nd About 68% of the scores should fall between Part 2 What percentage of scores would be above 628 ts of the scores would be above 628

Explanation / Answer

#4.
According to Chebyshev's theorem, at least 11/k2 of the data lie within k standard deviations of the mean.

Part 1
69 - 4*8 = 37 and 69 + 4*8 = 101
1-1/4^2 = 0.9375
At least 93.75% of the values fall between 37 and 101

Part 2
69 - 3*8 = 42 and 69 + 3*8 = 93
1-1/3^2 = 0.8889
At least 88.89% of the values fall between 45 and 96

#5.
mean = 4*60 = 240
sd = 31
mean - 4*sd = 240 - 4*31 = 116
mean + 4*sd = 240 + 4*31 = 364

At least 93.75% of the data will lie between 116 and 364 minutes

#6.
mean = 249200
sd = 47400
mean - 4*sd = 249200 - 4*47400 = 59600
mean + 4*sd = 249200 + 4*47400 = 438800

At least 93.75% of the data will lie between $59,600 and $438,800

#7.
mean = 508 and sd = 40
Part 1
mean - 1*sd = 508 - 40 = 468
mean + 1*sd = 508 + 40 = 548

About 68% of the scores should fall between 468 and 548

Part 2
(628 - 508)/40 = 3
(100-99.7)/2 = 0.15%

0.15% of the scores would be above 628

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