Suppose that quiz scores in a beginning statistics class have a mean of 6.6 with a standard deviation of 0.6. Using Chebyshev's Theorem, what is the minimum percentage of quiz scores between 4.8 and 8.4

88.89%

Step-by-step explanation:

Calculation for what is the minimum percentage of quiz scores between 4.8 and 8.4

The calculated minimum percentage value of quiz scores between

First step is to calculate the lower k value using this formula

k=(x1−m)/s

Where:

x1 represent first random value 4.8

x2 represent second random value 8.4

m represent mean 6.6

s represent standard deviation 0.6

Let plug in the formula

k=(4.8−6.6)/0.6

k=−1.8/0.6

k=−3

Second step is to calculate the upper k value

k=(x2−m)/s

k=(8.4−6.6)/0.6

k=1.8/0.6

k=3

Now let calculate the minimum percentage

Using this formula

Minimum percentage=1−1/k^2

Let plug in the formula

Minimum percentage=1−1/3^2

Minimum percentage=1−1/9

Minimum percentage=1−0.1111

Minimum percentage=0.8889*100

Minimum percentage=88.89%

Therefore the minimum percentage of quiz scores between 4.8 and 8.4 is 88.89%

75%

Step-by-step explanation:

5.2−6.40.6=−2  and 7.6−6.40.6=2.

This means that 5.2 lies 2 standard deviations below the mean and 7.6 lies 2 standard deviations above the mean.

According to Chebyshev's Theorem, at least 75% of data values lie within 2 standard deviations of the mean.

Therefore, we can say that 75% is the minimum percentage of quiz scores between 5.2 and 7.6.

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13