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Rashid [163]
3 years ago
15

A doctor's office has a computer file showing the heights of 100 male patients of one of the doctors of the practice. the height

s range from 18 inches to 68 inches. by accident, the greatest height has been recorded as 680 inches, instead of 68 inches. (a) does this affect the average? if so, by how much? (b) does this affect the median? if so by how much?
Mathematics
1 answer:
kumpel [21]3 years ago
7 0
A) Yes. we don't have enough information so this can not be determined.
B) No, it would just be an outlier.
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2/5 and 1/4 yw hope it helps

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Two random variables x and y are independent if the value of x does not affect the value of y. If the variables are not​ indepen
jeyben [28]

Answer: \mu_{x+y} = 3026

              \mu_{x-y}= 30

Step-by-step explanation: Average sum of the female and male's test score is the sum of expected value of each gender:

\mu_{x+y}=\mu_{x}+\mu_{y}

Assuming x represents the random male selected and y represents the random female selected:

\mu_{x+y}=1528+1498

\mu_{x+y}= 3026

The average sum of their scores is 3026.

Average difference is the difference between the expected value (mean) of each gender:

\mu_{x-y}=\mu_{x}-\mu_{y}

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The average difference of their scores is 30.

6 0
3 years ago
The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it
brilliants [131]

Answer:

(a) 11.25 and 1.68  

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....

where,  n = number of trials (samples) taken = 15

             r = number of success

             p = probability of success which in our question is % of cases in

                  which customer problems are resolved on the same day, i.e.;75%

So, here X ~ Binom(n=15,p=0.75)

(a) Expected number of problems to be resolved today = E(X)

            E(X) = \mu = n * p = 15 * 0.75 = 11.25

    Standard deviation = \sigma = \sqrt{n*p*(1-p)} = \sqrt{15*0.75*(1-0.75)} = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

     P(X = 10) = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}

                    = 3003*0.75^{10} *0.25^{5} = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

    = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}

    = 3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4} = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

    given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)  

= \binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}

= 1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}

= 0.6865

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Step by Step :
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