The number of customers who ate at a fast food restaurant between 1:00 P.M. And 3:00 P.M. Each day last week are listed here: 55, 37, 52, 50, 33, 46, 53 What's the standard deviation of the data set?

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ANSWER

The standard deviation is [tex]7.8[/tex] correct the nearest tenth.

EXPLANATION

The data set given to us is [tex]55,37,52,50,33,46,53[/tex].


The standard deviation of the given data set can be calculated using the formula,

[tex]Standard\:deviation=\sqrt{\frac{(x-\bar X)^2}{n} }[/tex]


where [tex]\bar X=\frac{\sum x}{n}[/tex]


[tex]\bar X=\frac{55+37+52+50+33+46+53}{7}[/tex]


[tex]\bar X=\frac{326}{7}[/tex]


[tex]\bar X=46.57[/tex]


We now find the standard deviation as follows;


[tex]SD=\sqrt{\frac{ (55-46.57)^2+(37-46.57)^2+(52-46.57)^2+(50-46.57)^2+(33-46.57)^2+(46-46.57)^2+(53-46.57)^2 }{7} }[/tex]


[tex]Standard\:deviation=\sqrt{\frac{71.06+91.58+29.48+11.76+184.14+0.32+41.34}{7} }[/tex]


[tex]Standard\:deviation=\sqrt{\frac{429.68}{7} }[/tex]


[tex]Standard\:deviation=\sqrt{61.38}[/tex]


[tex]Standard\:deviation=7.84[/tex]






Answer:

Standard deviation of the data set = 7.84

Step-by-step explanation:

To calculate the standard deviation of the data set first we take the mean of the data .

Mean = [tex]\frac{(55+37+52+50+33+46+53)}{7}[/tex]

         = [tex]\frac{326}{7}[/tex]

         = 46.57

Now we subtract the mean from data set and square the answer.

55 - 46.57 = 8.43² = 71.06

37 - 46.57 = -9.57² = 91.58

52 - 46.57 = 5.43² = 29.48

50 - 46.57 = 3.43² = 11.76

33 - 46.57 = -13.57² = 184.14

46 - 46.57 = -0.57² = 0.32

53 - 46.57 = 6.43² = 41.34

Then we will take the mean of squared result.

[tex]\frac{(71.06+91.58+29.48+11.76+184.14+0.32+41.34)}{7}[/tex]

=  [tex]\frc{429.68}{7}[/tex]

Variance = 61.38775509

Lastly take the square root of variance

= [tex]\sqrt{61.38775509}[/tex]

= 7.835033828 rounded to 7.84

Standard deviation of the data set = 7.84