by a neemal distribution with a tire of this SThe treed life of a particular bra
ID: 2947788 • Letter: B
Question
by a neemal distribution with a tire of this SThe treed life of a particular brand of tine is a rmarble best brand will last longer than 57,100 miles meanof s0,000 miles an astandand deviation of 2900 miles What is the probablity a particular t D) 0.7266 A)0.1587 )0.8413 C) 0.2266 ind the Indicateds-score 7)Determine the two Hores that separate the middle96% of the distribution fromthe area in touldthe standard normal distribution A) -2.05 and 2.05 B) 0 and 2.05 C)-175 and 1.75 D) -233 and 2.33 Find the value of za 8) 2005 A) 1.75 8) 0.52 C)-1.645 D) 1.645 Find the indicated percentile. 9) Assume that the random variable X is normally distributed with mean 50 and standard deviation ?-12. Find the 24th percentile for X. A) 58.52 B) 42.92 C) 38.6 D) 41.48 Provide an appropriate response. 10) The tread life of a particular berand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 3000 miles. What warranty should the company use if they want 96% of the tires to outlast the warranty? A) 63000 mi B) 54,750 mi C) 57,000 mi D)65250 ? Use a normal probability plot to asses whether the sample data could have come from a population that is normally distributed 11) Determine whether the following normal probability plot indicates that the sample data could have come from a population that is normally distributed. A) not normally distributed B) normally distributed Provide an appropriate response. 12) If sample data are taken from a population that is normally distributed, a normal probability plot of the observed data values versus the expected z scores will A) be approximately linear C) have no discernable pattern B) have a correlation coefficient near O D) look exponential in natuneExplanation / Answer
Solution6:
mean=60000
sd=2900
P(X>57100)
z=x-mean/sd
=57100-60000/2900
Z=-1
P(Z>-1)
=P(Z<1)
=0.8413
ANSWER:0.8413
OPTION B
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