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mr Goodwill [35]
3 years ago
14

About 7% of the population has a particular genetic mutation. 400 people are randomly selected. Find the standard deviation for

the number of people with the genetic mutation in such groups of 400.
Mathematics
1 answer:
Deffense [45]3 years ago
5 0

Answer:

5.10 people

Step-by-step explanation:

Calculation to Find the standard deviation for the number of people with the genetic mutation

Using this formula

Standard deviation = √ ( n * p * q)

Where,

n = 400

p = .07

q = 100%-7% = 93% or .93

Let plug in the formula

Standard deviation =√400*07*.93

Standard deviation =√26.04

Standard deviation = 5.10 people

Therefore the standard deviation for the number of people with the genetic mutation is 5.10 people.

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The science club has raised $154. 24 to buy food for a polar bear cub at the zoo. It costs $53.55 to feed the cub each week. How
fomenos

Answer:

$59.96

Step-by-step explanation:

Given:

The science club has raised $154. 24 to buy food for a polar bear cub at the zoo.

It costs $53.55 to feed the cub each week.

Question asked:

How much more money does the club need to raise to feed the cub for one month (four weeks)?

Solution:

Cost of feed for the cub for 1 week = $53.55

Cost of feed for the cub for 4 weeks = $53.55 \times 4 = $214.20

Money raised by the science club = $154. 24

More money needed by science club = Cost of feed for 4 weeks - Money raised by the science club

More money needed by science club = $214.20 - $154. 24

                                                               = $59.96

Thus, the science club will be needed $59.96 more money  for one month (four weeks) feed.

                                           

7 0
3 years ago
one possible combination of homeroom groups is 18 groups of 10 students and 5 groups of 12 students. which equation uses the x =
vfiekz [6]

Answer: y - 5 = -5/6 (x - 18)

Step-by-step explanation:

The point-slope form of a linear equation is written using the slope of the line and one point in the line. From part A, the slope of the line representing this situation is m = -5/6.

Since x represents the number of 10-student groups and y represents the number of 12-student groups, the combination of 18 groups of 10 students and 5 groups of 12 students is represented by the point (18,5).

6 0
3 years ago
Trevor tutors French for $12 an hour and scoops ice cream for $8 an hour. He is going to work 20 hours this week. At least how m
Arlecino [84]

Answer:

the answer is 16 hours

Step-by-step explanation:

i just multiplied 12(the amount he gets from tutoring) and multiplied it by different numbers  to see which one gets him past $190

8 0
3 years ago
Jeremy has 3 pounds of ground beef to make hamburgers. How many 13-pound hamburgers can he make? 12 9 6 1
Law Incorporation [45]

Answer: 9

Step-by-step explanation:

6 0
3 years ago
Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

ln(0.01) \geq -\lambda+ln(1+\lambda)

\lambda-ln(1+\lambda)+ln(0.01) \geq 0

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
3 years ago
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