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Please help with the following statistics question, There is some evidence to su

ID: 3377096 • Letter: P

Question

Please help with the following statistics question,

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. Demonstrate that X = the number of workers that belong to a union is a binomial random variable. Find the following probabilities: At most three California workers belong to a union. At least four but at most seven California workers belong to a union. Exactly eight California workers belong to a union. Less than twenty California workers do not belong to a union. More than nine but at most fourteen California workers do not belong to a union. What is the mean number of California workers that do belong to a union? Use the correct notation. What is the standard deviation for this probability distribution? Use the correct notation. Find P(mu - 2sigma

Explanation / Answer

a)

X is a binomial random variable as it only has two possible outcomes, that is, "belonging to a union" and "not belonging to a union". It also has a constant probability of success, 0.184.

Also, it seems here that the persons choose independently whether to join a union or not. Hence, X is a binomial random variable.

b)

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    26      
p = the probability of a success =    0.184      
x = the maximum number of successes =    3      
          
Then the cumulative probability is          
          
P(at most   3   ) =    0.269050207 [answer]

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c)

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    4      
x2 =    7      
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    26      
p = the probability of a success =    0.184      
          
Then          
          
P(at most    3   ) =    0.269050207
P(at most    7   ) =    0.910147397
          
Thus,          
          
P(between x1 and x2) =    0.64109719   [answer]

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d)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    26      
p = the probability of a success =    0.184      
x = the number of successes =    8      
          
Thus, the probability is          
          
P (    8   ) =    0.052810822 [answer]
  

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