Automobiles traveling on a road with a posted speed limit of 65 miles per hour are checked for speed by a state police radar system. The following is a frequency distribution of speeds. Speed (miles per hour) Frequency 45 up to 55 70 55 up to 65 360 65 up to 75 250 75 up to 85 110 (See the Excel Data File.) The mean speed of the automobiles traveling on this road is the closest to ________.

Answers

Answer 1
Answer:

Answer:

The mean speed of the automobiles traveling on this road is the closest to 65 mph.

Step-by-step explanation:

frequency distribution of speeds.

Speed (mph) | Frequency

45 up to 55 | 70

55 up to 65 | 360

65 up to 75 | 250

75 up to 85 | 110

Using the midpoint method, we represent each group/class of speeds with the midpoint speed, then go ahead to compute the mean.

Let the speed be x

The frequency be f

x | f

50 | 70

60 | 360

70 | 250

80 | 110

Mean = (Σfx)/(Σf)

Σfx = (50×70) + (60×360) + (70×250) + (80×110) = 51,400

Σf = 70 + 360 + 250 + 110 = 790

Mean = (Σfx)/(Σf)

Mean = (51400/790) = 65.06 mph ≈ 65 mph

The mean speed of the automobiles traveling on this road is the closest to 65 mph

Hope this Helps!!!


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Answers

Answer:

The 90% confidence interval is  0.199 <  p < 0.261

The sample size to develop a 95% confidence interval is n = 2032  

Step-by-step explanation:

From the question we are told that

   The sample size is n =500

    The sample proportion is  \^ p = 0.23

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of  (\alpha )/(2) is  

   Z_{(\alpha )/(2) } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{(\alpha )/(2) } * \sqrt{(\^ p (1- \^ p))/(n) }

=>   E =  1.645 * \sqrt{(0.23 (1- 0.23))/(500) }

=>   E =  0.03096

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.23  -0.03096  <  p < 0.23  +  0.03096

=>   0.199 <  p < 0.261

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  (\alpha )/(2) is  

   Z_{(\alpha )/(2) } =  1.96

The margin of error is given as E = 0.01

Generally the sample size is mathematically represented as  

    n = [\frac{Z_{(\alpha )/(2) }}{E} ]^2 * \^ p (1 - \^ p )

=>    n = [(1.96 )/(0.01) ]^2 *0.23 (1 - 0.23 )      

=>    n = 2032  

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Answers

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Answers

Answer:

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Step-by-step explanation:

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Answers

We are given number in words " two hundred fifty three thousandths".

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Answers

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Step-by-step explanation:

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