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 67^th percentile compare with the mean diameter?
A) 0.201 inch below the mean
B) 0.132 inch below the mean
C) 0.132 inch above the mean
D) 0.201 inch above the mean
E) 0.440 inch above the mean

Answer 1

Answer:

The correct answer is C.

Step-by-step explanation:

$67^{th}$ percentile is represented as the value of probability in the norma distribution.

We look for a z-value with a left-tail of 0.67.

z=0.44

where,

$z=\frac{x-M}{SD}$

M: mean

SD: standar deviation

x=z×SD+M=0.44×0.3+M=M+0.132

The diameter of an orange at the 67^th percentile is 0.132 inch above the mean.