Find the density function of y = ez , where z ∼ n(µ, σ2). this is called the lognormal density, since log y is normally distributed.

Y = e^2 -> ln y = z [e^2 > 0 -> y > 0]

Fy (y) = P (Y ≤ y)

= P (e^2 £ y)

= P (Z ≤ ln y)

= Fx (ln y)

Differentiating, fy (y) = 1/y fz (ln y)

= 1/y * [(1 / σ sqrt of 2π) e^(1/2 ((ln y - µ)/a)^2)] , y > 0

Which is the required density function of y = e^2

pooopoopopoopop

step-by-step explanati∵∵∴∴on:

36 girls

step-by-step explanation:

the 48 boys is 4 parts of the ratio

divide 48 by 4 to obtain one part of the ratio

= 12 ← 1 part of the ratio

3 parts = 3 × 12 = 36 ← number of girls