Let the mean and the variance of 5 observations x _ 1 , x _ 2 , x _...

Mean $\overset{¯}{x} = \dfrac{∑ {x}_{i}}{5} = \dfrac{24}{5} ⇒ ∑ {x}_{i} = 24 . . . \left(i\right)$

Variance ${\sigma}^{2} = \dfrac{∑ {x}_{i}^{2}}{5} - {\left(\dfrac{24}{5}\right)}^{2} = \dfrac{194}{25}$

$⇒ ∑ {x}_{i}^{2} = 154 . . . \left(ii\right)$

Also given mean$\dfrac{{x}_{1} + {x}_{2} + {x}_{3} + {x}_{4}}{4} = \dfrac{7}{2} ⇒ {x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} = 14$

$⇒ {x}_{5} = 10$ (from \left(i\right)$)</p><p>and variance${\sigma}^{2} = \dfrac{{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2}}{4} - \dfrac{49}{4} = a$</p><p>$⇒ {x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2} = 4a + 49$</p><p>$⇒ {x}_{5}^{2} = 154 - 4a - 49$(from \left(ii\right)$)

$⇒ 100 = 105 - 4a ⇒ 4a = 5$

Hence $4a + {x}_{5} = 15$