Cans of regular coke are labeled as containing 12 oz. statistics students weighted the content of 5 randomly chosen cans, and found the mean weight to be 12.13. assume that cans of coke are filled so that the actual amounts are normally distributed with a mean of 12.00 oz and a standard deviation of 0.12 oz. find the probability that a sample of 5 cans will have a mean amount of at least 12.13 oz.

0.0078

Step-by-step explanation:

We assume that cans of Coke are filled so that the actual amounts are normally distributed with a mean of 12.00 oz and a standard deviation of 0.12 oz.

As per given , we have

 and

Sample size : n= 5

Let be the sample mean amount of coke in can.

Then, the  probability that a sample of 5 cans will have a mean amount of at least 12.13 oz will be :_

 [By using z-table]

Hence, the required probability = 0.0078

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752