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A study investigated about 3000 meals ordered from Chipotle restaurants using the online site Grubhub. Researchers calculated
the sodium content (in milligrams) for each order based on Chipotle's published nutrition information. The distribution of
sodium content is approximately normal with mean 2000 mg and standard deviation 500 mg.
(b) What percent of the meals ordered had between 1200 mg and 1800 mg of sodium?
Round your answer to 4 decimal places and then convert to a percentage.

Respuesta :

s1m1

Answer:

≈28.62%

Step-by-step explanation:

Calculate number of standard deviations are 1,200 and 1,800 from the mean 2000.  

z= (x-mean)/ standard deviation

z= (1,200-2000)/ 500 = -1.6

z= (1,800-2000)/ 500 = -.4

Find the percentage to the left of z that is -1.6 and -.4 standard deviations away from the mean by reading standard normal table for z-score.

z = -1.6 is .05840 so 5.84%

z= -.4 is .34458 so 34.458%

Find the percentage between 1,200mg and 1,800mg:

34.458 - 5.84 = 28.618%

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