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on a particular stretch of highway, the state police know that the average speed is 62 mph with a standard deviation of 5 mph. on a busy holiday weekend, the police are concerned that people travel too fast. so they randomly monitor speeds of a sample of 50 cars and record an average speed of 66 mph. let's assume, for the moment, that people travel at the same speed on holiday weekends as they typically do on other days. calculate the mean and standard deviation of the sampling distribution of sample means for samples of size n

Respuesta :

The probability of a sample of 50 cars recording an average speed of 66 mph or higher is 0.

What is a standard deviation?

Data dispersion in regard to the mean is quantified by a standard deviation. Data are said to be more closely grouped around the mean when the standard deviation is low and more dispersed when the standard deviation is high.

Here, we have

µ = 62 , σ = 5

P ( X < 66 )

By applying the standard deviation formula, we get

Z = ( X - µ ) / (σ/√(n)

Z = ( 66 - 62 ) / ( 5 / √50 )

Z = 5.6569

P (  ( X - µ ) / ( σ/√(n)) < ( 66 - 62 ) / ( 5 / √(50) )

P ( X < 66 ) =  P ( Z < 5.66 )

P ( X < 66 ) = 1

P ( X > 66 )  = 1 - P ( X < 66 )

By applying the standard deviation formula, we get

Z = ( X - µ ) / ( σ / √(n))

Z = ( 66 - 62 ) / ( 5 / √ ( 50 ) )

Z = 5.6569

P (  ( X - µ ) / ( σ / √ (n)) > ( 66 - 62 ) / ( 5 / √(50) )

P ( Z > 5.66 )

P ( X > 66 )  = 1 - P ( Z < 5.66 )

P ( X> 66 )  = 1 - 1

P ( X> 66 )  =  0

Hence, the probability of a sample of 50 cars recording an average speed of 66 mph or higher is 0.

To learn more about the standard deviation from the given link

https://brainly.com/question/26941429

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