The number of accidents per day at a large factory is noted for each of 64 days,

Question

The number of accidents per day at a large factory is noted for each of 64 days, with a mean of 3.58. The population standard deviation for all accidents is known to be 1.52 accidents. With what confidence can we assert that the mean number of accidents per day at the factory is between 3.20 and 3.96 accidents? The number of absences at a school is studied and the population standard deviation of all absences is found to be 1.1. a. What sample size should be chosen to find the mean number of absences per month of school children to within plusminus 0.2 at a 90% confidence level? What sample size should be chosen to find the mean number of absences per month of school children to within A plusminus 0.2 at a 95% confidence level? Comparing the results from (a) and (b) above, what is your observation of how the confidence level changes with the sample size for the same interval? Use MATLAB's normcdf() function to solve the following problems. Please include a sketch for each problem along with your handwritten MATLAB script. a. Calculate the area (probability) from Z = -infinity to Z=1.96 Calculate the probability from Z = 0 to Z =2.4 Calculate the probability from Z = -1.96 to Z=1.96 5. Use MATLAB's norminv() function to solve the following problems. Please include a sketch for each problem along with your handwritten MATLAB script. Calculate the value of Z that corresponds to the bottom 2% Calculate the values of Z for a 93% Confidence interval

Explanation / Answer

2.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    3.2      
x2 = upper bound =    3.96      
u = mean =    3.58      
n = sample size =    64      
s = standard deviation =    1.52      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.954499736   [ANSWER]  

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