Women's heights are normally distributed with mean 63.5 in and standard deviation of 2.5 in. A social organization for tall people has a requirement that women must be at least 70 in tall. What percentage of women meet that requirement?

Question

anonymous · Answer

@campbell_st  help? :)

anonymous · Answer

anything? :/

campbell_st · Answer

ok you need to standardise the score

$$z = \frac{70 - 63.5}{2.5}$$

gives a z score of 2.6

next you need to use the z score with the standard normal curve

mathslover · Answer

First, ask yourself how far way is 70 from the mean of 63.6? It is equal to 70 - 63.6 = 6.4 inches.

Next, ask yourself how many standard deviations is this? Well, every standard deviation equals 2.5 inches, so there are 6.4 / 2.5 = 2.56 standard deviations. So, 70 is exactly 2.56 standard deviations to the right of the mean.

Finally, you can use the Standard Normal to find the answer P(X>70) since this is equivalent to:

P(z > 2.56) = 0.0052 or 0.52% of women will be taller than 70 inches.

Hope that helped

anonymous · Answer

0.9953? 99.53%?

anonymous · Answer

so, wait im confused.. is it 0.52?

campbell_st · Answer

to find the percentage greater than the minimum height its 

100 - P(z = 2.6)

P(z > 2.6) = 100 - 99.53 
                  = 0.47%

campbell_st · Answer

then on 0.47% of women in the population will meet the requirement