The height of 5 students -in cm- are as below- nbsp- 140- 145- 130- 135- 150 Their mean value is nbsp- nbsp-125 cm140 cm145 cm150 cm

The correct option is B 140 cmMean = Sum of all observations/Total number of observations nbsp; nbsp; nbsp;=140 +145+ 130+135+150/5 =700/5 = 140 ...

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Standard VIII

Mathematics

Mean for an Ungrouped Data

The height of...

Question

The height of 5 students (in cm) are as below.
{140, 145, 130, 135, 150}
Their mean value is .

125 cm

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140 cm

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145 cm

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150 cm

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Solution

The correct option is B 140 cm
$Mean = \frac{S u m o f a l l o b s e r v a t i o n s}{T o t a l n u m b e r o f o b s e r v a t i o n s}$

$= \frac{140 + 145 + 130 + 135 + 150}{5} = \frac{700}{5} = 140 c m$

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Similar questions

The table below shows the number of students in class 10 of a school, classified according to their heights:

Height

(cm)

Number of Students

120 − 125

125 − 130

130 − 135

135 − 140

140 − 145

145 − 150

150 − 155

155 − 160

Calculate the mean height

Q. Draw a histogram of the following data.

Height of student (cm)	135 - 140	140 - 145	145 - 150	150 - 155
No. of students	4	12	16	8

The height in centimetres of $30$ students of class $6$ in a school are given below: $138, 130, 132, 138, 140, 138, 140, 152, 132, 140, 155, 140, 132, 138, 140, 150, 138, 140, 150, 155, 150, 152, 155, 140, 138, 132, 155, 138, 140, 132$ If $150 c m$ is the standard height of the students of the class, how many are below this?

Q. The class mark of the class 130−150 is
(a) 130
(b) 135
(c) 140
(d) 145

Q. Heights in cm of 12 students are given as:

154, 158, 155, 160, 165, 145, 148, 150, 147, 151, 166, 152

Make a frequency table taking 145-150 as first class interval.

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The height of 5 students (in cm) are as below. {140, 145, 130, 135, 150} Their mean value is .

The correct option is B 140 cmMean=Sum of all observationsTotal number of observations =140+145+130+135+1505=7005=140 cm

The height of 5 students (in cm) are as below.
{140, 145, 130, 135, 150}
Their mean value is .

The correct option is B 140 cm
$Mean = \frac{S u m o f a l l o b s e r v a t i o n s}{T o t a l n u m b e r o f o b s e r v a t i o n s}$

$= \frac{140 + 145 + 130 + 135 + 150}{5} = \frac{700}{5} = 140 c m$