Cham
Lv 5

# Standard Deviation

Rating

(a) First of all, the class marks of the 6 classes are: 143, 148, 153, 158, 163 and 168 cm

Therefore the mean of the students' heights is:

(16 × 143 + 27 × 148 + 32 × 153 + 48 × 158 + 30 × 163 + 7 × 168)/160 = 155.2 cm

Also, the standard deviation is: √[(Σfihi2)/160] = 6.75 cm where i is from 1 to 6.

(b) So, within 1 standard deviation, the range of height is from 155.2 - 6.75 = 148.45 cm to 155.2 + 6.75 = 161.95 cm

Therefore, it includes all students within the ranges 151 - 155 cm and 156 - 160 cm.

Then, for the range 146 - 150 cm, the actual value of the height is given by 145.5 ≦ H < 149.5.

Assuming uniform distrubution, the no. of students in this range and being taller than 148.45 cm is 27 × [(149.5 - 148.45)/5] = 5.67

Similarly, fot no. of students in the range of 161 - 165 cm but shorter than 161.95 cm is 30 × [(161.95 - 160.5)/5] = 8.7

So total no. of students within the range specified in the question is 5.67 + 8.7 + 32 + 48 = 94.37 with percentage = 58.98 %.

Source(s): My Maths knowledge