The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. they would like the estimate to have a maximum error of 0.1 gallons. a previous study found that for an average family the variance is 5.76 gallons and the mean is 17.9 gallons per day. if they are using a 99% level of confidence, how large of a sample is required to estimate the mean usage of water?

19,652

Step-by-step explanation:

Error (E) = (t×sd)/√n

√n = (t×sd)/E

sd = √variance = √5.76 = 2.4, E = 0.1

Assuming the average family is a family of 4

n = 4, degree of freedom = n-1 = 4-1 = 3, t-value corresponding to 3 degrees of freedom and 99% confidence level is 5.841

√n = 5.841×2.4/0.1 = 140.184

√n = 140.184

n = 140.184^2 = 19651.55

n = 19,652

step-by-step explanation:

3024 rhe base is what up in worry about

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560