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Delicious77 [7]
3 years ago
8

1. A consumer testing laboratory measured the calories per hot dog in 20 brands of beef hot dogs. Here are the results: 186 181

176 149 184 190 158 139 175 148 152 111 141 153 190 157 131 149 135 132 BY HAND Find the five-number summary of this distribution. Be able to explain your procedure. Compute the range and interquartile range. Explain what these numbers tell you about the variability in calories in different brands of all-beef hot dogs. Would a beef hot dog with 175 calories be in the top quarter of the data
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
3 0

Answer:

Step-by-step explanation:

Hello!

The five-number summary of a data set includes:

1) Minimum value

2) 1st Quartile (C₁)

3) Median (2nd Quartile) (Me)

4) 3rd Quartile (C₃)

5) Maximum value

First step is to arrange the observations in asending order:

<em>111</em>, 131, 132, 135, 139, 141, 148, 149, 149, 152, 153, 157, 158, 175, 176, 181, 186, 190, <em>190</em>

From this you can easily detect the min and max od the data set:

Min value: 111

Max value: 190

To detect the quartiles you have to calculate their positions first.

For even samples, the position of each quartile is:

PosC₁= n/4= 20/4= 5

PosMe= n/2= 20/2= 10

PosC₃= (3*n)/4= (3*20)/4= 15

This means that the first quartile will be the 5th observation, the median will be the 10th observation and the third quartile will be the 15th observation:

C₁= 139

Me= 152

C₃= 176

The range of a data set is defined as de difference between the max value and the min value:

R= Max-Min= 190-111= 79

It gives you an idea of how dispersed the data is.

The IQR is the distance between quartile 3 and quartile 1:

IQR= C₃ - C₁= 176-139= 37

Is also a measure of dispersion, it shows how disperse the mid 50% of the data is (arround the Me)

Would a beef hot dog with 175 calories be in the top quarter of the data?

No, the top quarter of the data is determined by the third quartile, in this data set, the top quarter is the values of at least 176 calories. (X≥176)

I hope this helps!

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What is (-3) + (-8)​
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Answer:

\huge \boxed{-11}

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(-3)+(-8)

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4 years ago
Read 2 more answers
Together, Linda and Penny sold 492 boxes of Girl Scout cookies. Penny
boyakko [2]

Answer:

222

Step-by-step explanation:

Let L = number of boxes Linda sold.

Let P = number of boxes Penny sold.

The sold a total of 492 boxes.

L + P = 492

Penny sold 48 more boxes than Linda.

P = L + 48

Substitute P with L + 48 in the first equation and solve for L.

L + L + 48 = 492

2L + 48 = 492

2L = 444

L = 222

Answer: Linda sold 222 boxes.

Check:

Penny sold 48 more boxes, so she sold 222 + 48 = 270.

Add the numbers they sold: 222 + 270 = 492

The difference is 48, and the total is 492. Our answer is correct.

5 0
3 years ago
The median of Rae's data is 2.22 minutes, and the median of Doris' data is 2.14 minutes. The interquartile range for Rae is , an
JulsSmile [24]

Complete question :

Rae and Doris are training to swim a 200-meter freestyle race. The table lists their practice times during training camp.

Rae’s Times(minutes)-2.12, 2.01, 2.46, 2, 2.22, 2.31, 2.23

doris times(minutes)-2.32, 2.19, 2.26, 2.03, 2.11, 2.14, 2.07

The median of Raes data is(drop down box answers, 2,2.03,2.14,2.22)minutes,and the median of doris data is(drop down box answers, 2,2.03,2.14,2.22)minutes. The interquartile range for rae is(drop down box answers, 0.01,0.13,0.26,0.30), and the interquartile range for doris is(drop down box answers, 0.02,0.06,0.18,0.19). The two data sets overlap(drop down box answers,very little, a lot

Answer:

Rae's median = 2.22

Rae's IQR = 0.30

DORIS median = 2.14

Doris IQR = 0.19

Step-by-step explanation:

Given :

Rae's time :

Ordered data : 2, 2.01, 2.12, 2.22, 2.23, 2.31, 2.46

Median = 1/2(n+1)th term ; n = 7

Median = 1/2(8) = 4th term

Median = 2.22

Q1 = 1/4(8)th term

Q1 = 2nd term = 2.01

Q3 = 3/4(n + 1)th term

Q3 = 3/4(8)th term

Q3 = 6th term = 2.31

Interquartile range = Q3 - Q1 = 2.31 - 2.01 = 0.30

DORIS Data:

Ordered data: 2.03, 2.07, 2.11, 2.14, 2.19, 2.26, 2.32

Median = 1/2(n+1)th term ; n = 7

Median = 1/2(8) = 4th term

Median = 2.14

Q1 = 1/4(8)th term

Q1 = 2nd term = 2.07

Q3 = 3/4(n + 1)th term

Q3 = 3/4(8)th term

Q3 = 6th term = 2.26

Interquartile range = Q3 - Q1 = 2.26 - 2.07 = 0.19

6 0
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natulia [17]
<span>(a) This is a binomial experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
​(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
​(c) Fewer than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
​(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
</span>
6 0
3 years ago
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