Please help with part eh I am using a ti83 calculator so ple

Please help with part e-h. I am using a ti83 calculator, so please give the commands that you used so I can check the work!

There is some evidence to suggest that businesses are moving out of states where unions are prevalent. In California, 18.4% of all workers belong to a union. Suppose 26 workers from California are selected at random. 3. a. Demonstrate that X the number of workers that belong to a union is a binomial random variable. Find the following probabilities: b. At most three California workers belong to a union c. At least four but at most seven California workers belong to a union d. Exactly eight California workers belong to a union. Less than twenty California workers do not belong to a union. e. 2) e. More than nine but at most fourteen California workers do not belong to a union. f. What is the mean number of California workers that do belong to a union? Use the correct notation g. What is the standard deviation for this probability distribution? Use the correct notation h. Find P(-2

Solution

e)

Here, n = 26, p = 1 - 0.184 = 0.816, as we refer to those who are not members of a union.

Note that P(fewer than x) = P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    26      
p = the probability of a success =    0.816      
x = our critical value of successes =    20      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   19   ) =    0.18846503

[You can do this part by doing binomcdf(26,0.816,20)]
          
Which is also          
          
P(fewer than   20   ) =    0.18846503

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e) [There are two e\'s!]

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    10      
x2 =    14      
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    26      
p = the probability of a success =    0.816      
          
Then          
          
P(at most    9   ) =    1.7898E-07
[binomcdf(26,0.816,9)]

P(at most    14   ) =    0.001101006
[binomcdf(26,0.816,14)]          
Thus,          
          
P(between x1 and x2) =    0.001100827   [ANSWER]

  
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f)

mean = n p = 26*0.184 = 4.784 [answer]

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g)

standard deviation = sqrt(n p (1-p)) = sqrt(26*0.184*(1-0.184)) = 1.975789462 [ANSWER]

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h)

By empirical rule, within 2 standard deviations from the mean has probability 0.9545. [ANSWER]