Statistics Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

Mode of frequency distribution of series II is 46.

Answer: (d)

Since, intelligence of students is an attribute, arithmetic mean is not suitable method.

Answer: (d)

$(i + 1)x^i = (i + 1)x^i$ where i = 1, 2, 3, ------- n

$Σ↖{n}↙{i = 1}(i + 1)$ = 1.2 + 2.3 + 3.4 + 4.5 + -------meters

= $Σ↖{n}↙{n = 1} T_n$

= Σ n (n + 1)

= $Σn^2 + Σn$

= ${(n + 1)n(2n + 1)}/6 + {n(n + 1)}/2$

Mean = $1/n [{(n + 1) n (2n + 1)}/6 + {n(n + 1)}/2]$

= ${(n + 1)}/2 [{2n + 1}/3 + 1]$

= ${(n + 1)}/2 {(2n + 4)}/3$

= ${(n + 1)(n + 2)}/3$

So, option (d) is correct.

Answer: (a)

Number of Type A pencil = ${50}/1$ = 50

Number of Type B pencil = $x/{1.50}$

Number of Type C pencil = ${20}/2$ = 10

Average = ${\text"Total money spent"}/{\text"totalno.of pencil"}$ = 1.25

= ${x + 50 + 20}/{50 + 10 + {x/{1.50}}}$ = 1.25

= 70 + x = 1.25 $(60 + x/{1.50})$

70 + x = 75.00 + ${1.25}/{1.50}$x

x - ${125}/{150}$ x = 5

= ${25}/{150}$ x = 5

x = 30

Answer: (b)

Number of total bulbs = 100

∴ 7 + x + 40 + y + 10 + 2 = 100

⇒ x + y = 41 ... (i)

The median life is 20 h, so median interval will be (18-23).

Here, l = 18, $N/2$ = 50

c = 7 + x, f = 40, h = 5

∴ Median = l + ${(N/2 - C)}/f$ × h

⇒ 20 = 18 + ${(50 – 7 – x)}/{40}$ × 5

⇒ 2 = ${50 – 7 – x}/8$

⇒ 16 = 50 – 7 – x

⇒ x = 43 – 16

⇒ x = 27

Missing frequency 'y' is 14.