(a) what is the probability that a bolt selected at random has a diameter
(i) less than 89mm
The p(<89) is determined by using the formula Z= x - xmean / sx
Z=89-100/5
Z=-2.2
Identify the probability by using your graphing calculator
Press 2nd / Dist / 2 (normalcdf)
Then type in normalcdf(-1x10^99,-2.2)
This covers the shaded region in the normal distribution graph solution= .0139
(ii) between 89mm and 102mm
This is done the same way as (i), simply find the z-value for 102 and find the probability of selecting the shaded area.
Z= 102-100/5
Z=.0364
normalcdf(-2.2, .0364)= .5006
Constant uses puts the machine out of adjustment. Every week 100 bolts are tested to ensure that the machine is operating to its required level of performance. The results are summarized in the table below.
| Frequencies | Observed Frequencies |
| D < 90 | 29 |
| 90 < D < 100 | 33 |
| 100 < D < 105 | 18 |
| 105 < D < 110 | 9 |
| Totals | 100 |
(b)Assuming that the machine produces bolts in a normal distribution, determine the expected frequency.
To determine the expected frequency, use the same normcdf procedure as above. Since 90 is 2 standard deviations from the mean, the calculator entry will look as follows: normcdf(-1x10^99, -2)=.0227
This value is multiplied by 100 (the total) so the final answer is 2.27 The rest of the table is as follows:
| Frequencies | Observed Frequencies | Expected Frequencies |
| D < 90 | 29 | 13.6 |
| 90 < D < 100 | 33 | 34.1 |
| 100 < D < 105 | 18 | 34.1 |
| 105 < D < 110 | 9 | 13.6 |
| Totals | 100 | |
The Chi-Squared test for goodness of fit is used to determine the whether the sample fits the required distribution.
(c)(i) Write the null hypothesis and alternatve Hypothesis for the test.
Ho (null hypothesis)= the machine behaves according to normal distribution
Ha(alternative hypothesis)= the machine does not behave according to normal distribution
(ii) Calculate the chi-squared statistic
To calculate the chi-square value, enter the observed frequency into list 1 of your calculator, and the expected frequencey into list 2 of your calculator by using the following sequence:
Stat-Edit(1)-select list one and enter values using the formula of chi-square= sum of (( observed values-expected values )^2)/expected values
Enter this sequence into your calculator: 2nd-list-math-sum, then enter (( L1-L2 )^2)/L2
The value = 60.2
(iii) Determine, to a 5% significance level, the machine needs to be adjusted. Justify you answer.
To determine the meaning to a 5% significance, use a chi-square table. Use the column labeled 0.95 Identify the degrees of freedom with the formula ( #of rows-1 ) x ( #of columns-1 ), in this case it equals 4
Because the chi-square value exceeds the table value (9.488) the null hypothesis is rejected. This means we accept the alternative hypothesis, and the machine does not need to be fixed.
