25 men from Pinellas County were randomly drawn from a population of 100,000 men and weighed. The average weight of a man from the sample was found to be 150 pounds with a standard deviation of 54 pounds, Assuming the survey follows a normal distribution, find the 95% confidence interval for the true mean weight of men.

95% confidence interval for the true mean weight of men is [127.71 pounds , 172.30 pounds].

Step-by-step explanation:

We are given that the average weight of a man from the sample was found to be 150 pounds with a standard deviation of 54 pounds

25 men from Pinellas County were randomly drawn from a population of 100,000 men.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

                             P.Q. =  ~

where, = sample average weight of a man = 150 pounds

            s  = sample standard deviation = 54 pounds

            n = sample of men = 25

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the true mean, is ;

P(-2.064 < < 2.064) = 0.95  {As the critical value of t at 24 degree of

                                        freedom are -2.064 & 2.064 with P = 2.5%}  

P(-2.064 < < 2.064) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

                                            = [ , ]

                                            = [127.71 pounds , 172.30 pounds]

Therefore, 95% confidence interval for the true mean weight of men is [127.71 pounds , 172.30 pounds].

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!