Need Help please and Thanks ALGEBRA 2-Alternate Exam 6-continued The tables belo

Question

Need Help please and Thanks ALGEBRA 2-Alternate Exam 6-continued The tables below give the ten most popular names given to girls borm in the U.S. in the years 1974 and 1999 and the percentages of girls given those names. Consider a group of 20 girls born in 1974, and another group of 20 born in 1999. Use the data in the tables to do problems 22-25 regarding these groups 26 Kimberly ·43% 1.43% #1 Jennifer #2 I Am 4,03% I 89% #7 | Melissa #8 | Lisa #9 | Stephanie #101 Rebecca 1.42% 1.26% 1.08% 0.97% #3 Michelle | 165% #4 Heather #5 | Angela 1.48% 1.46% #1 Emily #2 | Hannah #3 | Alexis #4 | Sarah #51 Samantha 1.36% 1.11% 0.99% 0.98% 0.98% #6 | Ashley #7 | Madison #81 Taylor 093% 0.93% 10.87% #91 Jessica #101 Elizabeth 0.84% 0.79% 22·The name Jennifer was approximately 3 times as popular in 1974 as Emily was in 1999, despite each being the #1 name in their respective years. Determine by calculation if you are 3 times as likely to have more than one Jennifer in the 1974 group of 20 as you are to have multiple Emilys in the 1999 groyp of 20. 4.48 0.0403 |-20% (0.gp3)"G.. a oy03)n_zS1(o c C-0.0403 ICDC

Explanation / Answer

For multiple, or at least 2 Jennifer's:

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    20      
p = the probability of a success =    0.0403      
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.955313118
          
Thus, the probability of at least   2   successes is  
          
P(at least   2 Jennifer's) =    0.044686882

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For multiple, or at least 2 Emily's:

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    20      
p = the probability of a success =    0.0136      
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.997588701
          
Thus, the probability of at least   2   successes is  
          
P(at least   2 Emily's   ) =    0.002411299

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Thus, the ratio of probabilities is

P(at least 2 Jennifers)/P(at lest 2 Emilys) = 18.53 [ANSWER]

Thus, it is 18.53 times more likely to have multiple Jennifers in a group of 20 than multiple Emilys in a group of 20.

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