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Question

Heights of 10 year old children, regardless of sex, closely follow a normal distribution with mean 56 inches and standard deviation 5.2 inches. Round answers to 4 decimal places. 

a) What is the probability that a randomly chosen 10 year old child is less than 50.4 inches? 

b) What is the probability that a randomly chosen 10 year old child is more than 62.3 inches? 

c) What proportion of 10 year old children are between 51.9 and 59.3 inches tall? 

d) 95% of all 10 year old child are less than  inches.

Answer 1

Hi Lakotah,

All of these questions require either a z-score table or calculator, and all require the classic equation in elementary statistics:

z= (x-mu)/sigma

x= desired value

mu= mean = 56

sigma= standard deviation = 5.2

a. You were asked for P(X<50.4), so:

z= (x-mu)/sigma

z= (50.4 - 56)/5.2

Compute z and look it up in the z-table (you may need to round to 2 decimal places). Report the corresponding probability as your solution.

b. Same procedure as a, but with one key difference at the end of the calculation. Here, x= 62.3, so:

z= (x-mu)/sigma

z= (62.3- 56)/5.2

Again, compute z, and again, look it up in the z table. This time, though, note that the question asked for greater than. Probabilities in z-table are for less than, so we need to use the Complement Rule aka the "One Minus Trick," which states:

P(Z>a)= 1-P(Z<a); a=any real number

Long story short; subtract the probability you find in the z-table from 1 to get your final answer.

c. Any time you are asked for a range of probabilities like this-- a<x<b for a normal (z) distribution--that means you will need to subtract the smaller probability from the larger probability. So we need to compute two z-scores here, which I will call z₁ and z_2.

z₁= (x₁-mu)/sigma

z₁= (51.9 - 56)/5.2

z₂= (x₂-mu)/sigma

z₂ = (59.3 - 56)/5.2

Compute those z-scores, look up both probabilities in the z-table, subtract the smaller from the larger, and you are done.

d. For this, we use the 68-95-99.7 Rule aka the Empirical Rule. The rule states that in normal distributions, 68% of data fall within one standard deviation of the mean, 95% fall within two, and 99.7% fall within three. This goes for both directions i.e. positive and negative. Here, we are interested in 95%, so two standard deviations from the mean, and we are interested specifically in the positive end. So:

x= mu + 2*sigma

x= 56 + (2*5.2)

Compute that, and you will have the 95th percentile for height in inches. I hope this helps.