Which shows only a vertical translation from the shaded figure onto the unshaded figure?

2. A teacher asked his class of 20 students, "What is your age? Their responses are shown in the dot plot below.

Karolina [17]

Answer:

Yes, average does represent the centre of this data.

Average = 15.9

Step-by-step explanation:

As this question is not complete and not correctly written. Let's try to find an answer for this question.

Total number of Students = 20 = n

and we are given this data:

14, 15, 17, 18, 16.

which is just five entries and total number of students are 20. 15 entries are missing in this questions. So, let's add our own values into this question.

New set of entries for 20 students.

1. 14

2. 15

3. 17

4. 18

5. 16

6. 19

7. 17

8. 16

9. 17

10. 17

11. 15

12. 20

13. 14

14. 17

15. 18

16. 19

17. 13

18. 20

19. 17

20. 16

So, let's calculate the average of these numbers.

Average = Sum of all entries/ Total number of entries.

Average = 318/20

Average = 15.9

So, the average age of all the students of class is 15.9.

Let's calculate median:

For median , data must be arranged in a order from least to greatest number. Median is a number from that data which is at half way. In case of odd numbers, it is easy to identify median because there is only one number. and in case of even numbers, median is the average of the two middle numbers. So, here we have even numbers and median is the average of two middle numbers = 17 + 17/2 = 17

Median = 17

Yes, average does represent the centre of this data.

8 0

3 years ago

Read 2 more answers

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)

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