A simple random sample from a population with a normal distribution of 103 body temperatures has x overbarequals98.90degrees Upper F and sequals0.62degrees Upper F. Construct a 95% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

The 95% confidence interval would be given by (96.98;100.82)    

Step-by-step explanation:

Information given

represent the sample mean

population mean (variable of interest)

s=0.62 represent the sample standard deviation

n=103 represent the sample size  

Confidence interval

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

  (1)

The degrees of freedom are given by:

Since the Confidence is 0.95 or 95%, the significance is and , and the critical value for this case would be

And replacing we got:

The 95% confidence interval would be given by (96.98;100.82)    

let soccer = x and golf = y

part a

1st equation: x + y = 125

2nd equation: x = y +45

part b

replace x in the 1st equation with the 2nd one:

y + 45 + y = 125

simplify:

2y + 45 = 125

subtract 45 from each side:

2y = 80

divide both sides by 2:

y = 80/2

y = 40

she plays golf for 40 minutes.

part c

no because she plays soccer for 45 minutes longer than golf 40 +45 =   85 minutes, not 80 minutes.

what was there favorite food