Question

The distribution of the diameters of a particular variety of oranges is approximately normal with a standard deviation of 0.3 inch. How does the diameter of an orange at the 67th percentile compare with the mean diameter?

Answer 1

Answer:

$x = 0.132 + \mu$

is the required relation of the diameter of an orange at the 67th percentile compare with the mean diameter.        

Step-by-step explanation:

We are given the following information in the question:

Standard Deviation, σ = 0.3 inch

We are given that the distribution of  diameters is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.67.

$P( X \leq x) = P( z \leq \displaystyle\frac{x - \mu}{0.3})=0.67$  

Calculation the value from standard normal z table, we have,  

$P( z \leq 0.440) = 0.67$

$\displaystyle\frac{x - \mu}{0.3} = 0.440\\x - \mu = 0.132\\ x = 0.132 + \mu$

which is the required relation of the diameter of an orange at the 67th percentile compare with the mean diameter.