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jek_recluse [69]
4 years ago
8

Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of op

eration, and low cost. Suppose the maximum speed of a moped is normally distributed with mean value 46.7 km/h and standard deviation 1.75 km/h. Consider randomly selecting a single such moped. (a) What is the probability that maximum speed is at most 50 km/h? (Round your answer to four decimal places.) (b) What is the probability that maximum speed is at least 47 km/h? (Round your answer to four decimal places.) (c) What is the probability that maximum speed differs from the mean value by at most 1.5 standard deviations? (Round your answer to four decimal places.)
Mathematics
1 answer:
Mashcka [7]4 years ago
3 0

Answer:

a) P(X<50)=0.9827

b) P(X>47)=0.4321

c) P(-1.5<z<1.5)=0.8664

Step-by-step explanation:

We will calculate the probability based on a random sample of one moped out of the population, normally distributed with mean 46.7 and standard deviation 1.75.

a) This means we have to calculate P(x<50).

We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:

z=\dfrac{X-\mu}{\sigma}=\dfrac{50-46.7}{1.75}=\dfrac{3.7}{1.75}=2.114\\\\\\P(X

b) We have to calculatee P(x>47).

We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:

z=\dfrac{X-\mu}{\sigma}=\dfrac{47-46.7}{1.75}=\dfrac{0.3}{1.75}=0.171\\\\\\P(X>47)=P(z>0.171)=0.4321

c) If the value differs 1.5 standard deviations from the mean value, we have a z-score of z=1.5

z=\dfrac{(\mu+1.5\sigma)-\mu}{\sigma}=\dfrac{1.5\sigma}{\sigma}=1.5

So the probability that maximum speed differs from the mean value by at most 1.5 standard deviations is P(-1.5<z<1.5):

P(-1.5

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Answer:

<h2> 1.69</h2>

Step-by-step explanation:

Let the five courses with their corresponding credit hours be represented as shown;

FST 201 = 4

FST 203 = 5

FST 112 = 1

FST 219 = 5

FST 223 = 4

If a student earned grades of​ B, B,​ A, C, and D respectively in those courses with the following grading system  A =​4, B =​3, C =​2, D =​1, and F =0, before we can get the student GPA, we need to know the total credit point for the five courses.

Credit point for each course = Number of credit hour * point for each grade

FST 201 = 4 * 4 = 16points (A)

FST 203 = 5 * 4 = 20points (A)

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Total credit point = 16+20+4+20+16 = 76points

Total credit point gotten by the student is calculated as thus;

FST 201 = 4 * 3 = 12points (B)

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FST 223 = 4 * 1 = 4points (D)

Total credit point gotten by the student = 45points.

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