Asurvey found that women's heights are normally distributed with mean 63.2 in. and standard deviation 2.4 in. the survey also found that men's heights are normally distributed with a mean 67.2 in. and standard deviation 2.7. complete parts a through c below. a. most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 2 in. find the percentage of women meeting the height requirement?

99.51%

Step-by-step explanation:

Given : A survey found that women's heights are normally distributed.

Population mean :

Standard deviation:

Minimum height = 4ft. 9 in.=

Maximum height = 6ft. 2 in.=

Let x be the random variable that represent the women's height.

z-score :

For x=57, we have

For x=74, we have

Now, by using the standard normal distribution table, we have

The probability of women meeting the height requirement :-

Hence, the percentage of women meeting the height requirement = 99.51%

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.