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Question
ex#28: mitchell is on his schools track team. he recorded his times in the 100 - yard dash at each track meet last year.
100 - yard dash time (sec.)
a) what is his slowest time?
b) what is his fastest time?
c) what percent of his time is faster than 24 seconds?
ex#29 angie babysits to earn extra spending money. she recorded how much money she earned each time she babysat for a year.
a) what is her median pay?
b) what is her highest pay?
c) what is her lowest pay?
d) what percent of the time does she earn less than $25?
pretest/practice for summative#3
unit 3/7/11 practice (data analysis and problem solving)
unit 4/8/12 advanced algebra (no quadratics since that was sum#2)
unit 5/9/13 geometry/trig stuff
unit 3/7/11 practice (data analysis and problem solving)
#1 probability:
number of contestants by score and day
| 5 out of 5 | 4 out of 5 | 3 out of 5 | 2 out of 5 | 1 out of 5 | 0 out of 5 | total | |
|---|---|---|---|---|---|---|---|
| day 2 | 2 | 3 | 5 | 5 | 4 | 1 | 20 |
| day 3 | 3 | 3 | 4 | 5 | 3 | 2 | 20 |
| total | 7 | 9 | 13 | 16 | 9 | 6 | 60 |
the same 20 contestants, on each of 3 days, answered 5 questions in order to win a prize. each contestant received 1 point for each correct answer. the number of contestants receiving a given score on each day is shown in the table above.
no contestant received the same score on two different days. if a contestant is selected at random, what is the probability that the selected contestant received a score of 5 on day 2 or day 3, given that the contestant received a score of 5 on one of the three days?
Ex - 27 (Daily tip totals)
Step1: Identify median from box - and - whisker plot
The line inside the box represents the median. For the daily tip totals box - and - whisker plot, the median is at $150$.
Step2: Identify highest value
The end of the right - hand whisker represents the highest value. Here, the highest tip is $180$.
Step3: Calculate range
The range is the difference between the highest and lowest values. The lowest value is $120$ (start of the left - hand whisker) and the highest is $180$. So, range = $180 - 120=60$.
Ex - 28 (100 - yard dash times)
Step1: Identify slowest time
The end of the right - hand whisker represents the slowest time. So, the slowest time is $28$ seconds.
Step2: Identify fastest time
The end of the left - hand whisker represents the fastest time. So, the fastest time is $18$ seconds.
Step3: Calculate percentage faster than 24 seconds
The box - and - whisker plot shows that the third quartile ($Q_3$) is at $24$ seconds. $25\%$ of the data lies above $Q_3$. So, the percentage of times faster than 24 seconds is $25\%$.
Ex - 29 (Babysitting pay)
Step1: Identify median pay
The line inside the box of the box - and - whisker plot represents the median. The median pay is $30$.
Step2: Identify highest pay
The end of the right - hand whisker represents the highest pay. The highest pay is $60$.
Step3: Identify lowest pay
The end of the left - hand whisker represents the lowest pay. The lowest pay is $10$.
Step4: Calculate percentage earning less than $25$
The first quartile ($Q_1$) is at $20$ and the median ($Q_2$) is at $30$. The value $25$ is between $Q_1$ and $Q_2$. Since $25\%$ of the data lies below $Q_1$ and we assume a uniform distribution between $Q_1$ and $Q_2$, the percentage of times she earns less than $25$ is approximately $25\%+( \frac{25 - 20}{30 - 20})\times25\%=25\% + 12.5\%=37.5\%$.
Probability problem
Step1: Find number of contestants who got a 5 on one of the three days
The number of contestants who got a 5 on Day 1 is 2, on Day 2 is 2, and on Day 3 is 3. So, the total number of contestants who got a 5 on one of the three days is $2 + 2+3 = 7$.
Step2: Find number of contestants who got a 5 on Day 2 or Day 3
The number of contestants who got a 5 on Day 2 is 2 and on Day 3 is 3. So, the number of contestants who got a 5 on Day 2 or Day 3 is $2 + 3=5$.
Step3: Calculate conditional probability
The conditional probability formula is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In terms of counts, if $A$ is the event of getting a 5 on Day 2 or Day 3 and $B$ is the event of getting a 5 on one of the three days, then the probability is $\frac{5}{7}$.
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Ex - 27:
a) $150$
b) $180$
c) $60$
Ex - 28:
a) $28$ seconds
b) $18$ seconds
c) $25\%$
Ex - 29:
a) $30$
b) $60$
c) $10$
d) $37.5\%$
Probability problem: $\frac{5}{7}$