This is a homework question that I don't know how to solve...

 

The heights of 16-month-old oak seedlings are normally distributed with a mean of 31.5 cm and a standard deviation of 10 cm.

What is the range of heights between which 5 % of the seedlings will grow?

Can you show me how to answer this using proper mathematical notations. I am not sure what that means.

30.873 cm and 32.127 cm

Step-by-step explanation

We have

Mean = ?μ? = 31.5 cm

Standard deviation = ?σ? = 10 cm

P(- z < X < z) = 5 % = 0.05

Since normal curve is symmetric about mean

P(0 < X < Z ) = 0.05/2 = 0.025

From standard normal distribution table , Z value for P(0 < X < z) = 0.025

z = 0.0627

P(- 0.0627 < X < 0.0627) = 5 % = 0.05

Also

Z value is given as

?z=σx−μ??

z = ?±? 0.0627

Range of heights is given as

?x=μ+zσ? = 31.5 + 0.0627x10 = 32.127 cm

?x=μ−zσ? = 31.5 - 0.0627x10 = 30.873 cm

Therefore between 30.873 cm and 32.127 cm heights  5 % of the seedlings will grow i.e

P(30.873 < x < 32.127) = 5 %