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algol [13]
3 years ago
7

The distribution of passenger vehicle speeds traveling on a certain freeway in California is nearly normal with a mean of 73.7 m

iles/hour and a standard deviation of 4.77 miles/hour. (a) What percent of passenger vehicles travel slower than 80 miles/hour? (Round your answer to two decimal places.) % (b) What percent of passenger vehicles travel between 60 and 80 miles/hour? (Round your answer to two decimal places.) % (c) How fast do the fastest 5% of passenger vehicles travel? (Round your answer to four decimal places.) mph (d) The speed limit on this stretch of the freeway is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the freeway. (Round your answer to two decimal places.)
Mathematics
1 answer:
Schach [20]3 years ago
4 0

Answer:

A) 90.32%

B) 90.11%

C) 81.546 miles per hour

D)77.93%

Step-by-step explanation:

Given data:

speed  = 73.7 miles/hour

standard deviation = 4.77 miles /hour

a) P(X< 80) = P(z< \frac{80 - 73.7}{4.77}

                   = P (z< 1.31)

                   = 0.90320  = 90.32%

b) P(60 < X< 80) = P(\frac{60 - 73.7}{4.77} < z< \frac{80 - 73.7}{4.77})

                   = P (-2.872< z< 1.31)

                   = P (z<1.31) - P(z<-2.87)

                   = 0.90320 - 0.00205

                   = 0.90115  = 90.11%

c) P(X <X) = 0.95

    z = 1.645

\frac{x - 73.7}{4.77} = 1.645

x = 81.546  miles per hour

d)P(X< 70) = P(z> \frac{70 - 73.7}{4.77})

                   = P (z> -0.775)

                   = 0. 0.7793  = 77.93%

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