Determine the number of standard deviations that includes all data values listed for the situation. The mean height of a tree in an orchard is 11.8 feet; the standard deviation is 1.43 feet. 12.5 ft, 9.8 ft, 13.5 ft, 11.2 ft, 12.3 ft, 14.2 ft, 11.7 ft, 9.8 ft, 12.6 ft, 10.4 ft

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Answer:

Therefore the number of standard deviation is 1.41 ft.

Step-by-step explanation:

Mean: The average of all numbers is known as mean.

[tex]Mean=\frac{\textrm{sum of all number }}{\textrm{total number} }[/tex]

To find out the number of standard deviation,

  • First we have find out the mean of the distribution.
  • Then subtract mean from each number.
  • Squares of the difference.
  • Mean of the squares.
  • square root of the mean.

Given mean of the orchard tree is 11.8.

Given distribution is 12.5 ft, 9.8 ft, 13.5 ft,11.2 ft, 12.3 ft , 14.2 ft, 11.7 ft, 9.8 ft , 12.6 ft, 10.4 ft.

In order to get the number of standard deviation, we have to subtract mean from each number .

Then (12.5-11.8)ft,(9.8-11.8)ft , (13.5-11.8)ft,(11.2-11.8) ft, (12.3-11.8) ft , (14.2 -11.8)ft, (11.7 -11.8)ft,( 9.8-11.8) ft , (12.6-11.8) ft,( 10.4-11.8) ft.

=0.7 ft, -2.0 ft,1.7 ft, -0.6 ft,0.5 ft,2.4 ft, -0.1 ft , -2.0 ft,0.8 ft,-1.2 ft

Therefore the number of deviation is

[tex]=\sqrt{\frac{(0.7)^2+(-2.0)^2+(1.7)^2+(-0.6)^2+(0.5)^2+(2.4)^2+(-0.1)^2+(-2.0)^2+(0.8)^2+(-1.2)^2}{10} }[/tex]

[tex]=\sqrt{1.984}[/tex]

= 1.4085

≈1.41

Therefore the number of standard deviation is 1.41 ft.