Answer:
0.9554,0.8188
Step-by-step explanation:
Given that a tire manufacturer estimates that their tires last, an average of 40,000 miles, with a standard deviation of 5000.
X - duration of tires is N(40000,5000)
Or Z = [tex]\frac{x-40000}{5000}[/tex] is N(0,1)
the probability that a randomly chosen tire from this manufacturer lasts between 30,000 miles and 50,000 miles
=[tex]P(30000<x<50000) \\= P(-2<Z<2)\\=0.9554[/tex]
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2) Given that the daily high temperature on October 31 in a certain city is normally distributed with 50 and 8
Y - daily high temp on Oct 31 is N(50,8)
Or Z = [tex]\frac{x-50}{8}[/tex] is N(0,1)
the probability that a randomly chosen tire from this manufacturer lasts between 46 degrees and 58 degrees
=[tex]P(46<x<58) \\= P(-2<Z<1)\\=0.4772+0.3416=0.8188[/tex]
