Mathematics, 29.02.2020 05:26, erbs2003

3.12 Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour.47 (a) What percent of passenger vehicles travel slower than 80 miles/hour? (b) What percent of passenger vehicles travel between 60 and 80 miles/hour? (c) How fast to do the fastest 5% of passenger vehicles travel? (d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

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answered: ashleyzamarripa08

a) 93.94% of passenger vehicles travel slower than 80 miles/hour

b) 93.53% of passenger vehicles travel between 60 and 80 miles/hour

c) The fastest 5% of passenger vehicles travel at speeds of 80.46 mph and higher.

d) 70.54% of the passenger vehicles travel above the speed limit on this stretch of the I-5.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 72.6, \sigma = 4.78$

(a) What percent of passenger vehicles travel slower than 80 miles/hour?

This is the pvalue of Z when X = 80. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{80 - 72.6}{4.78}$

Z = 1.55

Z = 1.55 has a pvalue of 0.9394

93.94% of passenger vehicles travel slower than 80 miles/hour

(b) What percent of passenger vehicles travel between 60 and 80 miles/hour?

pvalue of Z when X = 80 subtracted by the pvalue of Z when X = 60. So

X = 80

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{80 - 72.6}{4.78}$

Z = 1.55

Z = 1.55 has a pvalue of 0.9394

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 72.6}{4.78}$

Z = -2.64

Z = -2.64 has a pvalue of 0.0041

0.9394 - 0.0041 = 0.9353

93.53% of passenger vehicles travel between 60 and 80 miles/hour

At speed of X and higher, in which X is found when Z has a pvalue of 0.95. So X when Z = 1.645.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 72.6}{4.78}$

X - 72.6 = 1.645*4.78

X = 80.46

The fastest 5% of passenger vehicles travel at speeds of 80.46 mph and higher.

(d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

This is 1 subtracted by the pvalue of Z when X = 70. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{70 - 72.6}{4.78}$

Z = -0.54

Z = -0.54 has a pvalue of 0.2946

1 - 0.2946 = 0.7054

70.54% of the passenger vehicles travel above the speed limit on this stretch of the I-5.

answered: Guest

i dont know sorry

step-by-step explanation:

answered: Guest

The answer is : 63explanation 60 times 0.05360+363

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