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Chapter 2: Q 57. (page 140)

Watch the salt! Refer to Exercise 55About what percent of the meals

ordered contained between 1200 mg and 1800 mg of sodium?

Short Answer

Expert verified

Around 28.98% of the meals ordered contained between 1200 mg and 1800 mg of sodium.

Step by step solution

01

Given information

Daily allowance,x=1200mgor1800mg

Mean,μ=2000mg

Standard deviation,σ=500mg

02

Concept

The formula used:z=x−μσ

03

Calculation

Calculate the Z − score,

z=x−μσ=1200mg−2000mg500mg=−1.60

or

z=x−μσ=1800mg−2000mg500mg=−0.40

To find the equivalent probability, use the normal probability table in the appendix.

In the typical normal probability table forP(z<−1.60), look at the row that starts with -1.6 and the column that starts with.00.

Or

The usual normal probability table forP(z<−0.40)has a row that starts with -0.4 and a column that starts with.00.

P(1200<x<1800)=P(−1.60<z<−0.40)=P(z<−0.40)−P(z<−1.60)=0.3446−0.0548=0.2898=28.98%

Therefore,

Around 28.98% of the meals ordered contained between 1200 mg and 1800 mg of sodium.

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Most popular questions from this chapter

Normal curve Estimate the mean and standard deviation of the Normal density curve below.

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