NEED SERIOUS MATH STATISTICS HELP!?

I’m having a hard time in my stat class and don’t have any examples to go by. Please help me.

A tire manufacturer believes the tread life of its snow tires can be described by a normal distribution with a mean of 32,000 miles and a standard deviation of 2800 miles. Find the probability (rounding to tenth of percent) that a randomly selected tire will have a tread life:

a. that last at least 40,000 miles

b. that last less than 30,000 miles

c. that last between 30,000 and 35,000 miles

d. In planning a marketing strategy, a local tire dealer wants to offer a refund to any customer whose tires fail to last a certain number of miles. However, the dealer doesn’t want to take too big a financial risk with this guarantee. If the dealer is willing to give refunds to no more than 1 out of every 50 tires sold, for what mileage can he guarantee each tire to last ? (Round to the nearest whole number)

e. Find the probability that 150 randomly chosen tires will have a mean tread wear of greater than 35,000 miles ? (Round to tenth of a percent)

Can i use my ti 83 to solve. How do i know what formulas to use?

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2 Responses to NEED SERIOUS MATH STATISTICS HELP!?

  1. dominic b says:

    ill print this out and ask my teacher for the answer and i will get back to you asap Yes i am in the same class i don’t get it either :(

  2. godzyx says:

    [2nd][VARS]

    normalcdf (lower, upper, mean, SD)
    invNorm(percent, mean, SD)

    See attached links for an explanation of functions.

    a. normalcdf(40000, 1E99, 32000, 2800) = 0.2%
    b. normalcdf(-1e99, 30000, 32000, 2800) = 23.8%
    c. normalcdf(30000, 35000, 32000, 2800) = 62.0%
    d. 1/50 = 2%, so 98% invNorm(0.98, 32000, 2800) = 37750 miles
    e normalcdf(35000, 1E99, 32000, 2800/√(150)) = 0.0%

    The last one makes sense since there is only a normalcdf(35000, 1E99, 32000, 2800) = 14.2% chance a tire >35000 miles. So over 150 tires, there is no chance.

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