2. you have tested bearings for a particular machine. this resulted in a mean of 30,457,500 cycles with a standard deviation of 167,800 cycles (assume approximately normally distributed). (a.) what percentage of the bearings will last for 30,000,000 cycles? (b.) if the manufacturer is willing to accept a 5% failure rate, what warranty time should he put on the bearings?

(a) 99.68% of the bearings will last for 30,000,000 cycles

(b) The manufacturer should put 30,181,494 cycles of warranty time on the bearings

Step-by-step explanation:

Let X be the random variable that represents the cycles of the bearings. We know that X is approximately normally distributed with a mean of = 30, 457, 500 cycles and a standard deviation of = 167, 800 cycles. Then, we can assume that the normal probability density function for X is given by

(a) To answer this question, we should to compute the next probability , i.e., the probability that the cycles of a bearing is equal or greater than 30,000,000.

. So,

99.68% of the bearings will last for 30,000,000 cycles.

pnorm(30000000, mean = 30457500, sd = 167800, lower.tail = FALSE) (execute this in the R statistical programming language)

(b) We are looking for a value such that

and this value is = 30,181,494 cycles, i.e., we found the 5th percentile.

qnorm(0.05, mean = 30457500, sd = 167800) (execute this in the R statistical programming language)

You can use a table from a book to compute this figures or a programming language like the R statistical programming language.

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

how many players are there?

step-by-step explanation: