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A company that manufactures tires produces a tire that has average life span of 65,000 kilometers with a standard deviation of 5,200 kilometers. The distribution of the life spans of the tire is normal. What percent of the tires last at least 60,000 kilometers ?

Respuesta :

Answer:

=> P(X <= 62500) = P((X-mean)/sd < (62500-65000)/3000)

= P(Z < -0.8333)

= 1 - P(Z < 0.8333)

= 1 - 0.7967

= 0.2033

=> P(X > 68500) = P(Z > (68500-65000)/3000)

= P(Z > 1.1667)

= 0.1210

=> P(60500 < X < 69500) = P((60500-65000)/3000 < Z < (69500-65000)/3000 )

= P(-1.5 < Z < 1.5)

= 0.8664

=> for P(X < x) = 0.03, for Z = -1.88

X = mean + Z*Sd = 65000 - (1.88*3000) = 59360

Step-by-step explanation:

I'm not sure if that correct , I really try my best to help

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