6 (1 2 points) A certain gene typically appears in 20% of the A. What is the sta

Question

6 (1 2 points) A certain gene typically appears in 20% of the A. What is the standard deviation of the percent of people who hav together with -scores and your percentile table to a population. Use the Central Limit Theorem answer these questions the gene in samples of 400 people oB. What is the Z-Score for this data? C. What more ofthe people in a town of 1000 have the gene due to chance "lone? is the chance that 24% or 7. (12 point margin oferror of the poll is 4.5%. A. What is the confidence interval of the result? ts) A polling firm finds that 45% of the population of a state favors a wolf hunting season. The B. About how many people were polled? Show work to earn credit. 8. (12 points) The lengths of a particular species of insect are normally distributed with a mean of 5 mi and a standard deviation of 0.2 mm. What percent of the insects are OVER 5.4 mm?

Explanation / Answer

Question 6:

Let P = population propertion of certain gene typically appers in the population = 0.20

Q = 1-P= 0.80

p = sample propertion of certain gene typically appears.

E(p) = mean = P

and Standard deviation = sqrt(PQ/n) where n = sample size.

a) p = sample propertion of people who have the gene in the sample 400 people.

n = sample size = 400

Standard deviation (p) = Sqrt(PQ/n) = sqrt(0.20*0.80/400) = 0.02

S.D.(p) =0.02

b) Since sampling distribution of sample propertion is given by

p ~ N ( P , PQ/n)   

E(p) = P and S.D.(p) = sqrt(PQ/n).

By central limit theorem

Z = (p- E(P)) / sqrt ( PQ/n) ~ N(0,1)

Z is known Z-score for the data.

c) n = 1000

E(p) = 0.20 and S.D.(p) = sqrt(0.20*0.80/1000) = 0.0126

Required Probability = P( p>=0.24)

P ( (p- E(P)) / sqrt ( PQ/n) > = ( 0.24 -0.20)/0.0126)

= P( Z >= 3.1746)

from normal probability table

P( Z >= 3.1746) = 0.0008

Question 7:

P = population propertion of state favors a wolf hunting season. = 0.45

Margin of error = 0.045

a) By defination of confidence interval

the confidence interval for propertion is given by

( P - Margin of error , P + Margin of error)

= ( 0.45 -0.045, 0.45 +0.45)

= (0.405 , 0.495)

b) Margin of error = 0.045

Margin of error = Z alpha/2 * Sqrt(PQ/n)

where alpha =level of significance

For 95% confidence interval

Zalpha/2 = Z 0.025 = 1.96

1.96 * sqrt(0.45 *0.55/n) = 0.045

sqrt(0.2475/n) = 0.02295

squaring both sides

0.2475 /n = 0.000527

n = 0.2475/0.00052

n=475.96 =476 ( approximately)

n = 476

Question 8 :

X : Length of perticular species of insect.

X ~ N( 5, 0.22)

E(X) = 5 and S.d(X) = 0.2

By Central limit theorem

Z = (X-E(X))/S.d(X) ~ N(0,1)

Required Probability = P (X >5.4)

= P( (X-E(X))/S.d(X) > ( 5.4-5)/0.2)

=P( Z >2)

from normal probability table

P( Z >2) = 0.0228

Percent of the insects over 5.4 mm = 2.28%

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