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niku
Answers (1)
Mean weight of the packets = 1.24 kg Standard Deviation, SD = 0.04kg Variance = 0.04^2 = 0.0016 Standard Error, SE = 0.04/sqrt (25) • = 0.04/5 = 0.008 Considering 99.7% confidence level the means will lie between (1.2+3SE) and
