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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. About what percent of the meals ordered exceeded the recommended daily allowance of 2400 mg of sodium?

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Answer:

21.186%

Step-by-step explanation:

z = (x-μ)/σ,

where

x is the raw score = 2400mg

μ is the population mean = 2000mg

σ is the population standard deviation = 500mg

z = 2400 - 2000/500

z = 0.8

Probability value from Z-Table:

P(x<2400) = 0.78814

P(x>2400) = 1 - P(x<2400) = 0.21186

Converting to percentage:

0.21186 × 100

= 21.186%

Therefore, the percent of the meals

ordered that exceeded the recommended daily allowance of 2400 mg of sodium is 21.186%

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21.186% of the meals ordered exceeded the recommended daily allowance of 2400 mg of sodium.

It is known that 3000 meals ordered from Chipotle restaurants using the online site Grubhub.

And the distribution of sodium content is approximately Normal with mean 2000mg and standard deviation 500 mg.

We know,

[tex]z = \frac{(x- \mu)}{\(\sigma\)}[/tex],

And daily allowance of 2400 mg of sodium

x is the raw score = 2400mg

μ is the population mean = 2000mg

σ is the population standard deviation = 500mg

[tex]z = \frac{2400 - 2000}{500}[/tex]

z = 0.8

Probability value from Z-Table:

[tex]P(x<2400) = 0.78814\\P(x>2400) = 1 - P(x<2400) \\= 0.21186[/tex]

Converting to percentage:

[tex]= 0.21186 (100)\\= 21.186%[/tex]

Therefore, the percent of the meals  ordered that exceeded the recommended daily allowance of 2400 mg of sodium is 21.186%.

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