Question

1.

Answer 1

The probability of randomly selecting an order that has at least 760

calories is 0.16

Step-by-step explanation:

The number of calories in an order of poutine at a certain fast food

restaurant is approximately normally  distributed

1. The mean "μ" is 740 calories

2. The standard deviation "σ" is 20 calories

3. We need to find the probability of randomly selecting an order that

    has at least 760 calories (x ≥ 760)

At first let us calculate z-score

∵ z = (x - μ)/σ

∵ x = 760 calories

∵ μ = 740 calories

∵ σ = 20 calories

∴ z = $\frac{760-740}{20}=1$

Use the normal distribution table of z (area to the left of z-score) to

find the corresponding area to z-score of 1

∵ The area corresponding of z = 1 is 0.84134

∵ We interested in x ≥ 760, we need the area to the right of z-score

∴ P(x ≥ 760) = 1 - 0.84134

∴ P(x ≥ 760) = 0.15866 ≅ 0.16

The probability of randomly selecting an order that has at least 760

calories is 0.16

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