Question

svar - Evacuus vrv, uvnals. A charity receives 2025 contributions. Contributions are assumed to be independent and identically distributed with mean 3125 and standard deviations 25. Find the probability that the total contributions exceed 6,330,000. Ley 1100 donate the natural net weinde fundamenn EA LLC

Answer 1

By Central limit Theorem,

X_i (i=1,2,3...N) are independent and identically distributed with mean $\mu$ and standard deviation $\sigma$, then by Sum of X_i

> Y=
follows normal distribution with mean N$\mu$, and standard deviation $\sqrt{N}\sigma$

X_i: Contributions from i^th charity;

Y : Total Contribution :  $\sum_{i=1}^{N}X_{i}$

Given, N = 2025 , $\mu$=3125 and $\sigma$ = 25

$N\mu$ = 2025x3125 = 6328125

1125

No = 2025 x 25 =

Y follows normal with mean 6328125 and standard deviation 1125

Probability that the total contribution exceed 6,330,000 = P(Y>6330000)

P(Y>6330000) = 1 -P(Y$\leq$6330000)

Z-score for 6330000 = (6330000-6328125)/1125 = 1875/1125 = 1.67

From standard normal tables, P(Z$\leq$1.67) = 0.9525

P(Y$\leq$6330000) = P(Z$\leq$1.67) = 0.9525

P(Y>6330000) = 1 -P(Y$\leq$6330000) = 1 -0.9525 = 0.0475

Probability that the total contribution exceed 6,330,000 = P(Y>6330000) = 0.0475

Probability that the total contribution exceed 6,330,000 = 0.0475