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he Greensboro Convention & Visitors Bureau has determined that across all hotels in the Triad region, the price for a one-night stay follows a normal distribution with mean $143 and standard deviation $38. What is the probability that a randomly chosen hotel charges between $120 and $160 per night?

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Answer:

0.4002

Step-by-step explanation:

Given that :

Mean (m) = 143

Standard deviation (σ) = 38

What is the probability that a randomly chosen hotel charges between $120 and $160 per night?

(120 < x < 160)

Obtain the standardized score :

(x - m) / σ

x : 120

Zscore = (120 - 143) / 38 = −0.605263

p(Z < −0.605263) = 0.2725

x : 160

Zscore = (160 - 143) / 38 = 0.4473684

p(Z < 0.4473684) = 0.6727

p(Z < 0.4473684) - p(Z < −0.605263)

= 0.6727 - 0.2725

= 0.4002

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