The mean and standard deviation of 50 observations are 50 and 10. Later on, it was decided to omit an observation that was incorrectly recorded as 99…

Question

The mean and standard deviation of 50 observations are 50 and 10. Later on, it was decided to omit an observation that was incorrectly recorded as 99. If the variance of the remaining 49 observations is (λ)/49, then the value of λ is equal to

MARKS Content Team · Accepted Answer

Correct answer: 2550. x-=(∑x_1)/50=50 ⇒∑x_i=2500 Correct ∑x_i=2500-99=2401 Correct x-=2401/49=49 σ^2=(∑x)/i2n-x-^2 ⇒100=(∑x)/i250-2500 ⇒∑xi2=130000 Correct ∑xi2=130000-99^2=120199 Correct σ^2=(∑x)/i2n-x-^2 =120199/49-49^2 =(120199-117649)/49=2550/49=(λ)/49 Hence, λ=2550

The mean and standard deviation of $50$ observations are $50$ and $10 .$ Later on, it was decided to omit an observation that was incorrectly recorded as $99 .$ If the variance of the remaining $49$ observations is $\frac{λ}{49},$ then the value of $λ$ is equal to

Solution

More Statistics Questions

The mean and standard deviation of 50 observations are 50 and 10. Later on, it was decided to omit an observation that was incorrectly recorded as 99. If the variance of the remaining 49 observations is λ49, then the value of λ is equal to

Solution

More Statistics Questions

The mean and standard deviation of $50$ observations are $50$ and $10 .$ Later on, it was decided to omit an observation that was incorrectly recorded as $99 .$ If the variance of the remaining $49$ observations is $\frac{λ}{49},$ then the value of $λ$ is equal to