Data Management Normal Distribution Probability

Question

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?

i did

μ = 31.5 cm 

o = 10cm

5% = 95% - 90% 

Z - Score *Found using the z - score table*: 

95% = 1.64 and 90% = -1.64 

31.5 + 1.64(10) and 31.5 - 1.64(10) 

47.9 and 15.1

but someone else got 25.76 and 26.24. How do I get to that answer? Who's right?

Answer 1

you are right the way to find probability for the normal distribution is

z = (x - µ)/(σ)

so for the z score if we look at the normal distribution table we get z = 1.64 and z = -1.64

µ = 31.5 σ = 10

so plugging in the number we have the equaltions

1.64 = (x - 31.5) / (10)

-1.64 = (x - 31.5) / (10)

so when we solve each questions

x = 47.9

x = 15.1

Hope this helps