Question 3:
Mean ˉx=∑xN√p=∑x5∑x=5√pStandard deviation, σ=√∑x2N−ˉx22q=√1205−(√p)24q2=24−pp=24−4q2
The mean of five numbers is
√p
. The sum of the squares of the numbers is 120 and the standard deviation is 2q. Express p in terms of q.
Solution:Mean ˉx=∑xN√p=∑x5∑x=5√pStandard deviation, σ=√∑x2N−ˉx22q=√1205−(√p)24q2=24−pp=24−4q2
Question 4:
Solution:
Variance, σ2=∑x2N−ˉx26=12+42+p23−(1+4+p3)26=17+p23−(5+p3)26=17+p23−[25+10p+p29]6=51+3p2−25−10p−p292p2−10p+26=542p2−10p+28=0p2−5p+14=0(p−7)(p+2)=0p=−2 (not accepted)∴
A set of positive integers consists of 1, 4 and p. The variance for this set of integers is 6. Find the value of p.
Solution:
Variance, σ2=∑x2N−ˉx26=12+42+p23−(1+4+p3)26=17+p23−(5+p3)26=17+p23−[25+10p+p29]6=51+3p2−25−10p−p292p2−10p+26=542p2−10p+28=0p2−5p+14=0(p−7)(p+2)=0p=−2 (not accepted)∴