Asociologist wishes to estimate the average number of automobile thefts in a large city per day within two automobiles. he wishes to be 95% confident, and from a previous study the standard deviation was found to be 4.2. how kany days should he select to survey?

n=17

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

The margin of error is given by this formula:

   (a)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=17 rounded up to the nearest integer

answer: asdfghjkl,mnbv

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11