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Demand per hour for gasoline at a local station is normally distributed with a mean of 1510 gallons and std deviation of 57 gallons. What is the probability that demand for a particular hour is greater than 1600 gallons? Enter your answer as a decimal, rounded to 4 decimal places

Sagot :

fichoh

Answer:

0.057168

Step-by-step explanation:

Given :

Mean = 1510

Standard deviation = 57

P(X > 1600)

To obtain the Zscore we use

(x - mean) / σ

P(Z > 1600) = P(Z > (x - mean) / σ)

P(Z > (x - mean) / σ) = P(Z > (1600 - 1510) / 57)

P(Z > 1.579) = 0.057168 (Z probability calculator)

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