Respuesta :
Answer:
[tex]0.67(4)+ 36[/tex]
Option D
Step-by-step explanation:
Given that a producer ships boxes of produce to individual customers.(say x)
X is a normal random variable with mean = 36 and std dev =4 lbs
By definition of std normal variate we know that
[tex]\frac{x-36}{4}[/tex] is N(0,1)
75th percentile of std normal distribution is
0.675
Hence corresponding x would be
[tex]36+0.675(4)[/tex]
Option D matches with this value
Hence answer is option D
[tex]0.67(4)+ 36[/tex]
Answer:
Option D.
Step-by-step explanation:
Given information: The distribution of weights of shipped boxes is approximately normal.
Population mean = 36
Standard deviation = 4
We need to find the 75th percentile of the distribution
75% = 0.75
From z-table it is clear that the value of z at 0.75 is 0.674.
75th percentile = Mean + z value at 75%(Standard deviation)
75th percentile = 36 + 0.674(4)
In can be rewritten as
75th percentile = 0.67(4) + 36
Therefore, the correct option is D.