In: Statistics and Probability
The lengths of one-year-old baby girls, in a certain country, can be described using a Normal distribution with a mean of 29 inches and a standard deviation of 1.2 inches. Use this information to answer the below.
a. What is the probability a randomly chosen one-year girl, from this population, is between 27 and 31 inches tall?
b. How tall is a one-year old girl, from this population, who is in the 85th percentile?
c. What is the z-score (standardized score) for a 28 inch tall girl? .
d. Suppose a random sample of 20 one-year old girls is taken from this population. What is the probability that the average height is less than 29 inches?
Solution :
Given that ,
mean = = 29
standard deviation = = 1.2
a. P(27 < x < 31) = P((27 - 29)/ 1.2) < (x - ) / < (31 - 29) / 1.2) )
= P(-1.67 < z < 1.67)
= P(z < 1.67) - P(z < -1.67)
= 0.9525 - 0.0475
= 0.905
Probability = 0.905
b. Using standard normal table,
P(Z < z) = 85%
P(Z < 1.04) = 0.85
z = 1.04
Using z-score formula,
x = z * +
x = 1.04 * 1.2 + 29
x = 30.25
85th percentile =30.25
c. x = 28
z = x - /
= 28 - 29 / 1.2 = -0.83
z-score = -0.83
d. n = 20
= 29 and
= / n = 1.2 / 20 = 0.2683
P( < 29) = P(( - ) / < (29 - 29) / 0.2683)
= P(z < 0)
Using standard normal table,
= 0.5
Probability = 0.5