Question 549289
The weight of a new-born baby follows a normal distribution with mean 3.44 kg 
and standard deviation 1.31 kg. The probability of the average weight of n new- 
born babies being greater than 3.13 kg is 0.937. Find the value of n 
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x-bar = 3.13 
AND
z(3.13) = (3.13-3.44)/(1.31/sqrt(n)
Find the z-score with a left-tail of 1-0.937 = 0.0630:
invNorm(0.0630) = -1.5301
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So (3.13-3.44)/[1.31/sqrt(n)] = -1.5301
-0.31/(1.31/sqrt(n)) = -1.5301
1.31/sqrt(n) = -0.31/-1.5301 = 0.2026
sqrt(n) = 1.31/0.2026 = 6.4658
n = 41.8 rounding up to 49
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Cheers,
Stan H.
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