Probability of Height in Poor Countries: 94-112cms

[sumans]

In poor countries, the growth of children can be an important indicator of general levels of nutrition and health. Data from several studies suggest that a reasonable model for the probability distribution of the height (in cms) of a randomly selected 5 year old child is normally distributed with mean 100 cms and standard deviation 6 cms.

a) What proportion of height is between 94 and 112 cms?
b) What is the probability that a random chosen five year old child will be taller than 110 cms.
c) If two five-year old children are drawn at random, what is the probability that exactly one of them will be taller than 110 cms.

It is given that area under a standard normal distribution between z = 0 and given values of z are as follows:
For z = 1.00 area = 0.3413
For z = 1.67 area = 0.4523
For z = 2.00 area = 0.4772

 

[Valkarie]

For z = 2.33 area = 0.4983a) The proportion of height between 94 and 112 cm is the area under the normal curve between z = -2 and z = 2, which is equal to 0.4772 - 0.0228 = 0.4544. b) The probability that a random chosen five year old child will be taller than 110 cms is the area under the normal curve between z = 2 and z = infinity, which is equal to 0.4772 - 0.5 = 0.0228. c) The probability that exactly one of two randomly chosen five-year old children will be taller than 110 cms is the area under the normal curve between z = 2 and z = 2.33, which is equal to 0.4983 - 0.4772 = 0.0211.

 

[tacman]

a) The proportion of height between 94 and 112 cms can be found by calculating the z-scores for these values and then finding the area under the curve between these z-scores. The z-scores for 94 and 112 are -1 and +2 respectively. Using the given values for area under the curve, we can find the area between these z-scores as follows:

Area = 0.4772 - 0.3413 = 0.1359 or 13.59%

Therefore, approximately 13.59% of children in poor countries have a height between 94 and 112 cms.

b) To find the probability that a randomly chosen five year old child will be taller than 110 cms, we first need to find the z-score for 110 cms. This can be calculated as:

z = (110 - 100)/6 = 1.67

Using the given values for area under the curve, we can find the area to the right of this z-score as follows:

Area = 0.5000 - 0.4523 = 0.0477 or 4.77%

Therefore, there is a 4.77% probability that a randomly chosen five year old child will be taller than 110 cms.

c) The probability that exactly one of the two children will be taller than 110 cms can be calculated by finding the area under the curve for one child being taller than 110 cms and the other being shorter than 110 cms. This can be calculated as:

Area = (0.5000 - 0.4523) x (0.5000 - 0.4523) = 0.0228 or 2.28%

Therefore, there is a 2.28% probability that exactly one of the two children drawn at random will be taller than 110 cms.