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borishaifa [10]
3 years ago
15

The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000

miles and a standard deviation of 1000 miles. What warranty should the company use if they want 96% of the tires to outlast the warranty? 59,000 miles 58,250 miles 61,000 miles 61,750 miles
Mathematics
1 answer:
valina [46]3 years ago
3 0

Answer: 61,750 miles

Step-by-step explanation:

Given : The p-value of the tires to outlast the warranty = 0.96

The probability that corresponds to 0.96 from a Normal distribution table is 1.75.

Mean : \mu=60,000\text{ miles}

Standard deviation : \sigma=1000\text{ miles}

The formula for z-score is given by  : -

z=\dfrac{x-\mu}{\sigma}\\\\\Rightarrow\ 1.75=\dfrac{x-60000}{1000}\\\\\Rightarrow\ x-60000=1750\\\\\Rightarrow\ x=61750

Hence, the tread life of tire should be 61,750 miles if they want 96% of the tires to outlast the warranty.

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A man who moves to a new city sees that there are two routes he could take to work. A neighbor who has lived there a long time t
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Answer:

(a) Confidence Interval for Route A is (38.61, 41.39)min

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(b) No

Step-by-step explanation:

(a) Route A

Mean=40min, sd=3min, df=33.1, t=2.0675, n=20

CI = (mean + or - t×sd/√n)

CI = (40 + 2.0675×3/√20) = 40+1.39 = 41.39min

CI = (40 - 2.0675×3/√20) = 40 - 1.39 = 38.61min

CI for Route A is (38.61,41.39)min

Route B

Mean=43min, sd=2min, t=2.0675, n=20

CI = (43 + 2.0675×2/√20) = 43+ 0.93 = 43.93min

CI = (43 - 2.0675×2/√20) = 43 - 0.93 = 42.07min

CI for Route B is (42.07,43.93)min

(b) He should not believe the old timer's claim. The man saves an average of 3 minutes a day by driving Route A

3 0
3 years ago
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hram777 [196]

Answer:

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2x + y = 3
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