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1/6/06 12:25 AM
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bigbrawler
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Edited: 06-Jan-06
Member Since: 08/25/2003
Posts: 229
 
I have a math assignment for Applied math 30. Now if anyone could get back to me with some answers/exsplanations that be great. I have the first four questions solved could you tell me if they are right or not. If you could that would be great. The impact of a gene depends partly on whether it is domminant or reccessive. If a gene is dominant, then only one copy is required to produce an effect. Huntington disease results when one copy of the defective gene is inherited from either parent. 1) a couple is planning on to have children. One of the couple is heterozygous (Hh) for Huntington disease, and the other is homzygous (Hh) fof the normal gene. What is the probability that this couple will have a child who has huntington disease. (Answer 2\4) If this couple were to have three children then what would be the probability that none of these children would have huntington disease? (Answer 1/8) 2) A sample of 1007 people, each of whom has only one parent who is heterozygous for the Huntington disease fene, was taken. Determine the symmetric 95% confidence interval for the number of people in the sample who would have Huntington disease. (Answer 15.87) 3) A different couple both of whom are heterozygous for the Huntington disease gene, is planning to have children. What is the probability that this couple will have a child who has Huntington disease? 2/4 if this couple were to have three children, then what would be the probability that at least on of theese children would have Huntington disease? (Answer 56/64 or 7/8) 4) Although the dominant gene for Huntington disease is present at birth the symptoms of the disease usually do not appera until much later. The age at which symptoms appear is normally distributed with a mean of 44 years and a standard deviation of 10.3 years. What is the probability that symptoms of the disease will appear before age 35.
1/6/06 6:06 PM
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TapoutRookie
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Edited: 06-Jan-06
Member Since: 01/01/2001
Posts: 1733
You sure the first is correct? Is the dominant Gene the H? Cause both can not be Hh. They could be HH, Hh and or hh. I assumed that the dominat is Hh and the recessive is hh, which would be 1/4 and 3/4 for your answer instead.
1/11/06 5:42 PM
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pats0
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Edited: 11-Jan-06
Member Since: 09/14/2003
Posts: 69
For the first question, I think you ment to say one is heterozygous (Hh) for Huntington disease, and the other is homzygous (hh) fof the normal gene, but your answers are still correct since the combinations are Hh,Hh,hh,hh. I dont know #2. For number 3 the first answer is 3/4. since the 4 four possible combinations are HH,Hh,Hh,hh. For the second part, first u find out the probablity of all 3 not having Huntington disease, ie. 1/4*1/4*1/4, which is 0.015625. Then subtract that from 1 to get the probability of at least one having Huntington disease, which is 0.9844.
1/11/06 5:43 PM
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pats0
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Edited: 11-Jan-06
Member Since: 09/14/2003
Posts: 70
For #4, since the question involves normal distribution, you are looking for P(X less than 35). You first have to convert it into a z-score, ie. (35-44)/10.3 which becomes P(z less than -0.8738). This can be rewritten, due to the symmetry of normal distribution to 1-P(z less than 0.8738). Now to figure out P(z less than 0.8738) you need to look at a normal distribution chart or use a calculator. It's 1 - 0.8089 = 0.1911 Stupid forum won't let you use less than signs without a pro membership.
1/12/06 10:59 AM
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bigbrawler
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Edited: 12-Jan-06
Member Since: 08/25/2003
Posts: 254
Well thanks for the help guys I apreciate it alot.

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