The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25…

Let the observations be ${x}_{1},{x}_{2},{x}_{3},...,{x}_{8},45,50$x1​,x2​,x3​,...,x8​,45,50,

Mean$=50$=50

$$\Rightarrow \frac{{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{8}+45+50}{10}=50$$⇒10x1​+x2​+x3​+...+x8​+45+50​=50

$$\Rightarrow {x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{8}=405....(i)$$⇒x1​+x2​+x3​+...+x8​=405....(i)

Hence, new mean

$${(\bar{X})}_{\mathrm{new}}=\frac{405+20+25}{10}=45$$(Xˉ)new​=10405+20+25​=45

Now,

$$S.D=\sqrt{\frac{\sum _{i=1}^{8}{x}_{i}^{2}+{45}^{2}+{50}^{2}}{10}-{(50)}^{2}}$$S.D=10∑i=18​xi2​+452+502​−(50)2​

$$\Rightarrow 12=\sqrt{\frac{\sum _{i=1}^{8}{x}_{i}^{2}+4525}{10}-2500}$$⇒12=10∑i=18​xi2​+4525​−2500​

$$\Rightarrow \sum _{i=1}^{8}{x}_{i}^{2}=21915$$⇒i=1∑8​xi2​=21915

Now,

$${(\mathrm{Variance})}_{\mathrm{new}}=\frac{\sum _{i=1}^{8}{x}_{i}^{2}+{20}^{2}+{25}^{2}}{10}-{(45)}^{2}$$(Variance)new​=10∑i=18​xi2​+202+252​−(45)2

$$\Rightarrow {(\mathrm{Variance})}_{\mathrm{new}}=\frac{21915+{20}^{2}+{25}^{2}}{10}-{(45)}^{2}$$⇒(Variance)new​=1021915+202+252​−(45)2

$$\Rightarrow {(\mathrm{Variance})}_{\mathrm{new}}=2294-2025$$⇒(Variance)new​=2294−2025

$$\Rightarrow {(\mathrm{Variance})}_{\mathrm{new}}=269$$⇒(Variance)new​=269

Hence this is the correct answer.