In the united states, the mean average height of adult women is approximately 65.5 inches, with a standard deviation of 2.5 inches. if height is normally distributed, what percent of the women in this country are between 63 and 70.5 inches tall?

Z = (x - mean) / SD
The z-score for x = 63 is: z = (63 - 65.5) / 2.5 = -1
The z-score for x = 70.5 is: z = (70.5 - 65.5) / 2.5 = 2
Therefore we need the probability that -1 < z < 2. This is equivalent to the probability that z < -1, subtracted from the probability that z < 2. From z-tables, we find that this is equal to 0.9772 - 0.1587 = 0.8185. This means that 81.85% of women's heights fall within this range.

If it is my birthday, then it is snowing