normal distribution

[Raj Koothrapali]

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 75% of the seedlings will grow?
how do i do this?

please provide step by step detail

 

[Raj Koothrapali]

I know that you are supposed to create a range because its continuous data however when i convert the 75% to a z score and identify the value i get the wrong answer

i did (0.75-31.5) /10 = 3.075 this z score is too large to find on the table

 

[Raj Koothrapali]

I know that you are supposed to create a range because its continuous data however when i convert the 75% to a z score and identify the value i get the wrong answer

i did (0.75-31.5) /10 = 3.075 this z score is too large to find on the table , The main issue is that i need to convert two values into z score but dont seem to have enough info

i apologize if this doesnt make any sense, i'm very lost

 

[HallsofIvy]

I know that you are supposed to create a range because its continuous data however when i convert the 75% to a z score and identify the value i get the wrong answer

i did (0.75-31.5) /10 = 3.075 this z score is too large to find on the table

This not only is wrong- it makes no sense! You are told "The heights of 16-month-old oak seedlings are normally distributed with a mean of 31.5 cm " and asked "What is the range of heights between which 75% of the seedlings will grow? "

You cannot subtract 31.5 cm from 75%!

"75%" is not a height, it is "75% of all the seedlings" and is the probability. You are going the wrong direction- 0.75 is the probability and you want to find the corresponding x value. Rather than "(0.75- 31.5)/10" You want to find x so that (x- 31.5)/10 has probability 0.75.