Question

The average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 14. Suppose that 18 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution.

1. What is the distribution of XX? XX ~ N(,)
2. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
3. If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 69.1 and 73.
4. For the 18 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 69.1 and 73.
5. For part d), is the assumption that the distribution is normal necessary? YesNo

Answer 1

Solution :

Given that ,

mean = = 66

standard deviation = = 14

a.

X N (66 , 14)

b.

n = 18

= 66

= / n = 14 / 18 = 3.2998

N (66 , 3.2998)

c.

P(69.1 < x < 73) = P[(69.1 - 66)/ 14) < (x - ) /  < (73 - 66) / 14) ]

= P(0.22 < z < 0.5)

= P(z < 0.5) - P(z < 0.22)

= 0.6915 - 0.5871

= 0.1044

Probability = 0.1044

d.

= P[(69.1 - 66) /3.2998 < ( - ) / < (73 - 69.1) / 3.2998)]

= P(0.94 < Z < 1.18)

= P(Z < 1.18) - P(Z < 0.94)

= 0.881 - 0.8264

= 0.0546

Probability = 0.0546

e.

yes