Answer :
The probability that maximum speed differs from the mean value by at most 2.5 standard deviations is 0.8473
Let the required probability that the maximum speed of mopeds differs from the mean speed by 2.5 standard deviations be given by P(U±2.5).
Mean speed U =46.7km/hr
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
Thus the probability that maximum speed differs from mean value by at most 2.5 standard deviations is 0.8473
the question is incomplete and the full question is "Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of operation, and low cost. Suppose the maximum speed of a moped is normally distributed with a mean value of 46.7 km/h and a standard deviation of 1.75 km/h. Consider randomly selecting a single such moped
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"
problem on mean and standard deviation
brainly.com/question/16030790
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