Question

The number of different species of plants in some gardens is recorded.

Answer 1

(a) The mean of the number of different species of plants in some gardens is 4.5 .

(b) The standard deviation of the number of different species of plants in some gardens is 2.12 (Approximately)

It is given in the question that the number of different species recorded are:-

1 2 3 4 4 5 5 6 7 8

We have to find the mean and standard deviation of the given data of the number of plants of different species.

(a) We know that, the formula for mean is :-

Sum of all the observations/ Total number of observations

Hence,

Mean = $\frac{1+2+3+4+4+5+5+6+7+8}{10} =\frac{45}{10}=4.5$

(b) Standard deviation = $\sqrt{\frac{Sum(X-m)^{2} }{N} }$

Where,

X = each value

m = mean

N = number of values

Hence,

$Sum(X-m)^{2} =(1-4.5)^{2}+(2-4.5)^{2}+ (3-4.5)^{2}+ (4-4.5)^{2}+ (4-4.5)^{2}+ (5-5.5)^{2}+ (5-5.5)^{2}+ (6-5.5)^{2}+ (7-5.5)^{2}+ (8-5.5)^{2}\\ \\ Sum(X-m)^{2}=(-3.5)^{2} +(-2.5)^{2}+ (-1.5)^{2}+ (-0.5)^{2}+ (-0.5)^{2}+ (0.5)^{2}+ (0.5)^{2}+ (1.5)^{2}+ (2.5)^{2}+ (3.5)^{2}$

$Sum(X-m)^2=12.25+6.25+2.25+0.25+0.25+0.25+0.25+2.25+6.25+12.25\\\\=42.5$

Standard deviation = ${\sqrt{\frac{{42.25}}{10} }} =\frac{6.7}{\sqrt{10} }=\frac{6.7}{3.16}=2.12$  (approximately).

To learn more about mean, here:-

brainly.com/question/13451489

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