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mean = 10, SD = 4
P(x > 14) = P( z > 1) = normalcdf(1, 100) = 15.87%
= 4/4 = 1
.......
Sample 200..assuming the large sample has normal distribution
P(x < 6) = P(z < -1) = 15.87%
200(.1587) = 32 rounded UP
......
Area under the standard normal curve to the left of the particular z is P(z)
Note: z = 0 (represents the mean) 50% of the area under the curve is to the left and 50% to the right
