Summaries

A specialist teacher observes the time taken by each of the students with learning disabilities to complete a psychological task. She summaries the times using the following:

Time Taken(mins)	1-5	6-10	11-12	16-20
No. of Student	2	4	12	4

Using the given data, calculate the
(i) mean (3)
(ii) mode (4)
(iii) median (4)
(iv) range (2)
(v) variance (4)
(vi) standard deviation

(1) 5 2130 4 3140 1

Final Answer:

μ =  11.2727 min
m =  11.5 min
m₂ =  11.5 min
r =  19 min
s =  361.3636
σ =  19.0096

Step-by-step explanation:

$$\begin{gathered} a_1=(5+1)/2=3 \\ {a}_2=(10+6)/2=8 \\ {a}_3=(11+12)/2=\frac{23}{2}=11.5 \\ {a}_4=(16+20)/2=18 \\ \mu =\frac{2\cdot a_1+4\cdot a_2+12\cdot a_3+4\cdot a_4}{2+4+12+4}=\frac{2\cdot 3+4\cdot 8+12\cdot \frac{23}{2}+4\cdot 18}{2+4+12+4}=\frac{124}{11}\text{min}=11.2727\text{min} \end{gathered}$$

$m=\frac{11+12}{2}=\frac{23}{2}\text{min}=11.5\text{min}$

$$\begin{gathered} 3,3,8,8,8,8,11.5,11.5,11.5,11.5,11.5,11.5, \\ 11.5,11.5,11.5,11.5,11.5,11.5,18,18,18,18 \\ {m}_2=11.5=\frac{23}{2}\text{min}=11.5\text{min} \end{gathered}$$

$r=20-1=19\text{min}$

$s=2\cdot (3-\mu {)}^2+4\cdot (8-\mu {)}^2+12\cdot (11.5-\mu {)}^2+4\cdot (18-\mu {)}^2=2\cdot (3-\frac{124}{11}{)}^2+4\cdot (8-\frac{124}{11}{)}^2+12\cdot (11.5-\frac{124}{11}{)}^2+4\cdot (18-\frac{124}{11}{)}^2=\frac{3975}{11}=361\frac{4}{11}=361.3636$

$\sigma =\sqrt{s}=\sqrt{361.3636}=19.0096$

Try another example