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Vedmedyk [2.9K]
3 years ago
7

A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get

at least 4 donors every day to meet her quota. How many calls must she make to be 89.9% sure of meeting her quota?
Mathematics
1 answer:
Tems11 [23]3 years ago
8 0
An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899  (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
P(x)=C(n,x)p^x(1-p)^{n-x}where
p=probability of success
n=number of trials
x=number of successesC(n,x)=\frac{n!}{x!(n-x)!}

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more) 
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%.  The work is similar to that of n=10.
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VladimirAG [237]

Answer:

1st number: 20

2nd number: 25

3rd number: 75

Step-by-step explanation:

x + y + z = 120

x = y - 5

z = 3y

plug that in!

(y-5) + y + (3y) = 120

remove the parenthesis (I used them to show you what I was replacing), and add like terms.

5y - 5 = 120

move -5 to the other side

5y = 120 + 5

5y = 125

divide both sides by 5

y = 25

now that you have one number, use the equations from before to solve for the rest!

x = (25) - 5

z = 3(25)

_____________________________________________________

x = 20

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3 years ago
With a height of 68 ​in, Nelson was the shortest president of a particular club in the past century. The club presidents of the
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Answer:

a. The positive difference between Nelson's height and the population mean is: \\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).

c. Nelson's z-score: \\ z = -1.1739 \approx -1.174 (Nelson's height is <em>below</em> the population's mean 1.174 standard deviations units).

d. Nelson's height is <em>usual</em> since \\ -2 < -1.174 < 2.

Step-by-step explanation:

The key concept to answer this question is the z-score. A <em>z-score</em> "tells us" the distance from the population's mean of a raw score in <em>standard deviation</em> units. A <em>positive value</em> for a z-score indicates that the raw score is <em>above</em> the population mean, whereas a <em>negative value</em> tells us that the raw score is <em>below</em> the population mean. The formula to obtain this <em>z-score</em> is as follows:

\\ z = \frac{x - \mu}{\sigma} [1]

Where

\\ z is the <em>z-score</em>.

\\ \mu is the <em>population mean</em>.

\\ \sigma is the <em>population standard deviation</em>.

From the question, we have that:

  • Nelson's height is 68 in. In this case, the raw score is 68 in \\ x = 68 in.
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  • \\ \sigma = 2.3in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson​'s height and the​ mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

That is, <em>the positive difference is 2.7 in</em>.

b. How many standard deviations is that​ [the difference found in part​ (a)]?

To find how many <em>standard deviations</em> is that, we need to divide that difference by the <em>population standard deviation</em>. That is:

\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174

In words, the difference found in part (a) is 1.174 <em>standard deviations</em> from the mean. Notice that we are not taking into account here if the raw score, <em>x,</em> is <em>below</em> or <em>above</em> the mean.

c. Convert Nelson​'s height to a z score.

Using formula [1], we have

\\ z = \frac{x - \mu}{\sigma}

\\ z = \frac{68\;in - 70.7\;in}{2.3\;in}

\\ z = \frac{-2.7\;in}{2.3\;in}

\\ z = -1.1739 \approx -1.174

This z-score "tells us" that Nelson's height is <em>1.174 standard deviations</em> <em>below</em> the population mean (notice the negative symbol in the above result), i.e., Nelson's height is <em>below</em> the mean for heights in the club presidents of the past century 1.174 standard deviations units.

d. If we consider​ "usual" heights to be those that convert to z scores between minus2 and​ 2, is Nelson​'s height usual or​ unusual?

Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

\\ -2 < z_{Nelson} < 2

\\ -2 < -1.174 < 2

Then, Nelson's height is <em>usual</em> according to that statement.  

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Answer:

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