I really need help with these two questions!!!!!

Question

A class of 200 students takes an exam. The students’ scores are normally distributed with a mean of 82 and a standard deviation of 5.Find the number of students in the interval between 82 and 92. (Hints: What's the standard deviation here? Use the percentages in the normal distribution curve provided on the Formula Sheet. Remember: You're calculating the number of students.)

 

What is the likely range for the population proportion voting for Bernie Sanders, if 225 voters out of a sample of 947 voters say they are voting for Bernie Sanders?

Answer 1

mean = 82

1 std dev = 5

2 std dev = 10

 

for a normal distribution 

+/- 1 std dev = 68.2 %   or 34.1% to each side of the mean

+/- 2 std dev = 95.4 %   or 47.7% to each side of the mean

 

I hope this graphically shows the percents

 

                                 |

                    |<-----68.2%----->| 1 std dev

                                 |

         |<--------------95.4---------------->| 2 std dev

                                 |<----47.7%---->|

        72                    82                      92

 

so 47.7% of 200 is .477*200 = 95.4 

rounded to nearest integer = 95 students score from 82-92

 

 

population proportion of 225 out of 947 is just 225/947  = 0.2376  or 23.76%

 

 

In re-reading your problem,  if you need an estimate of the std deviation

of the pop proportion it is given by

 

SQRT( p*(1-p)/n )    where p est is 0.2376 (above)

 

so  sqrt(.2376(1-.2376)/n) = 0.0138  or 

+/- 1.38%  for 1 std dev.    or  0.0138*947  = 13.06  round to 13

+/- 2.76%  for 2 std dev     or  0.0276*947  = 26.12  round to 26

 

so 68.2 % prob  225 +/- 13    or   212-238  (68%)

and 95% prob  225 +/- 26            199-251  (95%)