The number of nuts in a can of mixed nuts is found to be normally distributed, with a mean of 500 nuts and a standard deviation
of 20 nuts. My can of mixed nuts has only 443 nuts. What is the z-score for this can of nuts? (Enter your answer to two decimal places.)
2 answers:
Answer: the z score is - 2.85
Step-by-step explanation:
The number of nuts in a can of mixed nuts is found to be normally distributed. The formula for normal distribution is expressed as
z = (x - u)/s
Where
x = number of nuts
u = mean number of nuts
s = standard deviation
From the information given,
u = 500 nuts
s = 20 nuts
x = 443
To determine z,
z = ( 443 - 500)/20 = - 2.85
Answer:
z = -2.85
Step-by-step explanation:
Since the number of nuts per can is normally distributed:
Mean number of nuts (μ)= 500 nuts
Standard Deviation (σ)= 20 nuts
X = 443 nuts
For any given number of nuts X, the z-score is given by:
The z-score for this can of nuts with 443 nuts is -2.85.
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