A researcher investigating mosquito populations recorded a list of estimates of mosquito populations at each of \(n\) different marshland locations in a certain country. The distribution of these estimates was symmetric about the average (arithmetic mean), and 70% of these population estimates were within one standard deviation of the mean. Of these estimates, 75 were more than one standard deviation below the mean. Approximately how many of the estimates were within one standard deviation of the mean?

A) 175
B) 350
C) 375
D) 425
E) 500
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Let M = Arithematic Mean
70% Population Estimate is within 1 Standard Deviation of Mean (1D)
Let this 70% = X
i.e. (M-1D) <= X <= (M+1D)

70% of estimate is within 1 SD of Mean, it implies that 30% of estimates lie in the region on the left side of (M-1D) and on the right side of (M+1D).

Since, the estimates are Symmetric around the mean,
15% of Estimate is on left side of (M-1D), and
15% of Estimate is on right side of (M+1D).

Given, 15% of Estimate on left side of (M-1D) = 75

Therefore, X can be determined by:
15/70 = 75/X
X = 350

Hence, Answer is B