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Oliga [24]
4 years ago
12

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 49 km/h? (Round your answer to four decimal places.)
(b) What is the probability that maximum speed is at least 48 km/h? (Round your answer to four decimal places.) 2451
(c) What is the probability that maximum speed differs from the mean value by at most 2.5 standard deviations? (Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question
Mathematics
1 answer:
Drupady [299]4 years ago
5 0

Answer:

a. 0.9049

b. 0.2297

c. 0.8473

Step-by-step explanation:

Given U = 46.7, SD = 1.75

and Z = (U - X)/sd

(a). We want to find P(X<=49)

P(X<=49) = P(Z<= 49-46.7/1.75)

= P(Z<= 1.31)

= 0.90490

~= 0.9049

(b) P(X>=48) = P(Z>=48-46.7/1.75)

=P(Z>=0.7429) = 1 - P(Z<=0.7429)

= 1 - 0.77035

= 0.22965

=0.2297

(c) The required probability is given by P(U±2.5).

P(46.7±2.5) = (44.2<= X <=49.2)

= P( 44.2<= Z <= 49.2)

= P(44.2 - 46.7/1.75 <= Z <= 49.2- 46.7/1.75)

= P( - 1.4286 <= Z <= 1.4285)

= P(1.4285) - P(-1.4285)

= P(1.43 ) - P(-1.43)

= 0.92364 - 0.07636

= 0.8473

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