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Assume that the speed of automobiles on an expressway, during rush hour is normally distributed with a mean of 59 mph in a standard divination of 2 mph what percentage of cars are traveling faster than 59 mph

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In this case 50% of the cars are traveling faster than 59 mph during rush hour.

We need to convert the speed of 59 mph into a standard normal variable (Z-score) to use the standard normal distribution table.

Z = (X - μ) /

σ = (59 mph - 59 mph) / 2 mph = 0

Find the area to the right: Since we want the percentage of cars faster than 59 mph, we need the area under the standard normal curve to the right of Z = 0.

Use a Z-table or calculator: Look up the area under the curve for Z > 0 in a Z-table or use a calculator function like NORMSDIST(0, 9999) in Excel, which gives you 50%.

Therefore, 50% of the cars are traveling faster than 59 mph during rush hour.

User Randall Stephens
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