I was wondering if you may help me with this question Suppos

I was wondering if you may help me with this question:

Suppose that x is a binomial random variable with n = 5, p = 0.41, and q = .59.

For each value of x, calculate p(x). (Round final answers to 4 decimal places.)

Find P(x 3). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

Find P(x < 3). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

Find P(x 4). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

Find P(x > 2). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

Calculate the interval [_x ± 2_x]. Use the probabilities of part b to find the probability that x will be in this interval. Hint: When calculating probability, round up the lower interval to next whole number and round down the upper interval to previous whole number. (Round your answers to 4 decimal places. A negative sign should be used instead of parentheses.)

Solution

PART B)

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 =    5      
p = the probability of a success =    0.41      
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.07149243
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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 =    5      
p = the probability of a success =    0.41      
x = the number of successes =    1      
          
Thus, the probability is          
          
P (    1   ) =    0.248405901

******************

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 =    5      
p = the probability of a success =    0.41      
x = the number of successes =    2      
          
Thus, the probability is          
          
P (    2   ) =    0.345242099

******************

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 =    5      
p = the probability of a success =    0.41      
x = the number of successes =    3      
          
Thus, the probability is          
          
P (    3   ) =    0.239914001

*****************

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 =    5      
p = the probability of a success =    0.41      
x = the number of successes =    4      
          
Thus, the probability is          
          
P (    4   ) =    0.08335995

***********************

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 =    5      
p = the probability of a success =    0.41      
x = the number of successes =    5      
          
Thus, the probability is          
          
P (    5   ) =    0.01158562

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