Question

2. Suppose body mass index (BMI) varies approximately to the normal distribution in a population of boys aged 2-20 years. A national survey analyzed the BMI for American adolescents in this age range and found the µ=17.8 and the σ=1.9. a) What is the 25th percentile of this distribution? (1 point) b) What is the z-score corresponding to finding a boy with at least a BMI of 19.27? (2 points) c) What is the probability of finding a boy with at least this BMI? (2 points)

Answer 1

Solution:

Given that,

mean = = 17.8

standard deviation = = 1.9

a ) the 25th percentile

Using standard normal table,

P(Z < z) = 25%

P(Z < z) = 0.25

P(Z < - 0.67) = 0.25

z = - 0.67

Using z-score formula,

x = z * +

x = - 0.67 * 1.9 + 17.8

= 16.527

The 25th percentile = 16.5

b )  The z-score

X = 19.27

Using z-score formula,

Z = X - /

Z = 19.27 -17.8 / 1.9

= 0.77

The z-score = 0.77

c ) ( X   19.27 )

= 1 - p ( x   19.27 )

= 1 - p ( x - /)   (19.27 - 17.8 /1.9)

= 1 - p( z   1.47 / 1.9 )

= 1 - p ( z 0.77 )   

Using z table

= 1 - 0.7793

=0.2207

Probability = 0.2207