Name: Variance and Standard Deviation for Ungrouped Distribution
Uploaded: 2019-07-19
Duration: 6 min 21 s
Description: In this video, let us learn to find the variance and standard deviation of ungrouped data. We will also learn the formula to find the discrete frequency dist...

Question

The temperatures, $T °$ Celsius, at seven locations in the Central Kalahari Game Reserve were recorded at $4$ p.m. one January afternoon. The values of $T,$ correct to $1$ decimal place, were:

$32.1, 31.7, 31.2, 31.5, 31.9, 32.2$ and $32.7 .$

Evaluate $Σ 10 (T - 30)$ and $Σ 100 {(T - 30)}^{2}$ and calculate standard deviation.

By $5 p . m .$ the temperature at each location had dropped by exactly $0.75 ° C .$ Find the variance of the temperatures at $5 p \cdot m .$

Accepted Answer

The temperatures, $T °$ Celsius, at seven locations in the Central Kalahari Game Reserve were recorded at $4$ p.m. one January afternoon.

$32.1, 31.7, 31.2, 31.5, 31.9, 32.2$ and $32.7 .$

We know that, from the property of $Mean (a x) = a \times Mean (x)$ , $Σ a x = a Σ x$ and $Σ a x^{2} = a Σ x^{2}$ .

Now, $Σ 10 (T - 30) = 10 \times ((32.1 - 30) + (32.7 - 30) + (31.2 - 30) + (31.5 - 30) + (31.9 - 30) + (32.2 - 30) + (32.7 - 30))$

$= 10 \times (2.1 + 1.7 + 1.2 + 1.5 + 1.9 + 2.2 + 2.7) = 133$

and

$Σ 100 {(T - 30)}^{2} = 100 \times ({(32.1 - 30)}^{2} + {(32.7 - 30)}^{2} + {(31.2 - 30)}^{2} + {(31.5 - 30)}^{2} + {(31.9 - 30)}^{2} + {(32.2 - 30)}^{2} + {(32.7 - 30)}^{2})$

$= 100 \times (2 . 1^{2} + 1 . 7^{2} + 1 . 2^{2} + 1 . 5^{2} + 1 . 9^{2} + 2 . 2^{2} + 2 . 7^{2}) = 2673$ .

We know that, $SD (a x) = a SD (x)$ , where $a$ is constant and $x$ is a random variable and formula of $SD (x) = \sqrt{\frac{\sum_{} x^{2}}{n} - {(\frac{\sum_{} x}{n})}^{2}}$ .

Here, $SD (10 x) = 10 SD (x)$ .......(i)

Thus, standard deviation is $\frac{1}{10} \sqrt{\frac{2673}{7} - {(\frac{133}{7})}^{2}} = 0.457 ° C$ .

We know that, $SD (a x + b) = a SD (x)$ , where $a, b$ are constants,

Here, $SD (10 x - 0.75) = 10 SD (x)$ ........(ii)

From (i) and (ii)

$\Rightarrow SD (10 x - 0.75) = SD (10 x)$

So, the standard deviation of the temperature at $5 pm$ is $0.4567$ .

We know that, variance $= {(Standard deviation)}^{2}$ .

Therefore, variance $= (0.45669 \dots)^{2} = 0.209 {({}^{\circ}C)}^{2}$ .

	1st April		1st June
	Mean	SD	Mean	SD
Bakery	$2.50$	$0.40$	$2.25$	$0.36$
Household	$5.00$	$1.20$	$4.75$	$1.25$
Tinned food	$4.00$	$0.60$	$4.10$	$0.60$
Fruit and veg	$2.00$	$0.40$	$2.00$	$0.50$

	1st April		1st June
	Mean	SD	Mean	SD
Bakery	$2.50$	$0.40$	$2.25$	$0.36$
Household	$5.00$	$1.20$	$4.75$	$1.25$
Tinned food	$4.00$	$0.60$	$4.10$	$0.60$
Fruit \& veg	$2.00$	$0.40$	$2.00$	$0.50$

	1st April		1st June
	Mean	SD	Mean	SD
Bakery	$2.50$	$0.40$	$2.25$	$0.36$
Household	$5.00$	$1.20$	$4.75$	$1.25$
Tinned food	$4.00$	$0.60$	$4.10$	$0.60$
Fruit & veg	$2.00$	$0.40$	$2.00$	$0.50$

Important Questions on Measures of Variation

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