33.293 ± 0.01= 33.303 and 33.383,
Step-by-step explanation:
We first need to fit a normal distribution , but neither the mean  nor the standard deviation is given . We therefore estimate the sample mean and sample standard deviation  s. Using the data we find ∑fx=378  and
∑fx²=1344  so that mean x` = 2.885 or 2.9  and standard deviation s =1.360
x     f         fx    x²     fx²
1 Â Â Â 27 Â Â Â Â Â Â 27 Â Â Â Â 1 Â Â Â Â Â 27
2 Â Â Â 30 Â Â Â Â Â 60 Â Â Â Â 4 Â Â Â Â Â 120
3 Â Â Â 29 Â Â Â Â Â 87 Â Â Â Â 9 Â Â Â Â Â 261
4 Â Â Â 21 Â Â Â Â Â Â 84 Â Â Â Â 16 Â Â Â Â 336
5 Â Â Â 24 Â Â Â Â Â 120 Â Â Â 25 Â Â Â Â 600 Â Â Â Â Â
   ∑f=131    ∑fx=378       ∑fx²=1344 Â
Mean = x`= ∑fx/  ∑f=  2.9
Standard Deviation = s= √∑fx²/∑f-(∑fx/∑f)²
         s= √1344/131 - (378/131)²
          s= √10.26-(2.9)²
           s= √10.26- 8.41
          s= √1.85= 1.360
Next we need to compute the expected frequencies for all classes and the value of chi square. The necessary calculations for expected frequencies , ei`s ( ei= npi`) where pi` is the estimate of pi together with the value of chi square are shown below.
Categories    zi`    P(Z<z)       pi`    Expected     Observed
                                  frequency ei   Frequency Oi
1 Â Â Â Â Â Â Â Â Â Â -1.39 Â Â Â 0.0823 Â Â 0.0823 Â Â Â 10.78 Â Â Â Â Â Â Â Â Â 27
2 Â Â Â Â Â Â Â Â Â -0.66 Â Â Â 0.2546 Â Â 0.1723 Â Â Â Â 22.57 Â Â Â Â Â Â Â Â 30
3 Â Â Â Â Â Â Â Â Â Â 0.07 Â Â Â Â 0.5279 Â Â Â 0.2733 Â Â Â 35.80 Â Â Â Â Â Â Â Â 29
4 Â Â Â Â Â Â Â Â Â 0.808 Â Â Â 0.7881 Â Â Â 0.2602 Â Â Â Â 34.08 Â Â Â Â Â Â Â Â 21
5 Â Â Â Â Â Â Â Â Â Â 1.54 Â Â Â Â 0.937 Â Â Â Â Â 0.1489 Â Â Â Â 19.51 Â Â Â Â Â Â Â Â 24
Next we need to compute the expected frequencies for all classes and the value of chi square. The necessary calculations for expected frequencies , ei`s ( ei= npi`) where pi` is the estimate of pi together with the value of chi square are shown below.
Categories      Expected     Observed    (oi-ei)²/ei
            frequency ei   Frequency Oi     OBSERVED VALUE
1 Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.78 Â Â Â Â Â Â Â Â Â 27 Â Â Â Â Â Â Â Â Â Â 24.41
2 Â Â Â Â Â Â Â Â Â Â Â Â Â 22.57 Â Â Â Â Â Â Â Â 30 Â Â Â Â Â Â Â Â Â Â Â Â 1.54
3 Â Â Â Â Â Â Â Â Â Â Â Â Â 35.80 Â Â Â Â Â Â Â Â 29 Â Â Â Â Â Â Â Â Â Â Â Â 1.29
4 Â Â Â Â Â Â Â Â Â Â Â Â 34.08 Â Â Â Â Â Â Â Â 21 Â Â Â Â Â Â Â Â Â Â Â Â Â 5.02
5 Â Â Â Â Â Â Â Â Â Â Â Â Â 19.51 Â Â Â Â Â Â Â Â 24 Â Â Â Â Â Â Â Â Â Â Â Â 1.033
Total                        131          33.293
There are five categories , we have used the sample mean and sample standard deviation , so the number of degrees of freedom is 5-1-2= 2
The critical region is chi square ≥ chi square (0.001)(2) =9.21
CONCLUSION:
Since the calculated value of chi square =9.21 Â does not fall in the critical region we are unable to reject our null hypothesis and conclude normal distribution provides a good fit for the given frequency distribution.