0.036179852 Round to nearest hundredth

The percentage of cotton in material used to manufacture men's shirts follows.

velikii [3]

Answer:

Median = 34.55

Quartiles:

Q1 = 33.1

Q2 = 34.55

Q3 = 35.6

Inter Quartile range = 2.5

Step-by-step explanation:

Step 1

We arrange the given the following data from lowest to the highest

32.0, 32.1, 32.1, 32.2, 32.4, 32.5, 32.5, 32.5, 32.8, 32.8, 32.8, 32.8, 32.9, 32.9, 33.0, 33.0, 33.2, 33.2, 33.2, 33.2, 33.5, 33.6, 33.8, 33.8, 33.9, 34.0, 34.0, 34.1, 34.1, 34.3, 34.4, 34.5, 34.6, 34.6, 34.6, 34.7, 34.7, 35.0, 35.0, 35.1, 35.1, 35.2, 35.4, 35.5, 35.5, 35.5, 35.5, 35.6, 35.6, 36.1, 36.3, 36.4, 36.4, 36.4, 36.5, 36.6, 36.7, 36.7, 37.0, 37.1, 37.4, 37.4, 37.6, 37.8

a. Construct a stem-and-leaf display for the data.

Stem and leaf Display

Stem | Leaf

32 | 0,1,1,2,4,5,5,5,8,8,8,8,9,9

33 | 0,0,2,2,2,2,5,6,8,8,9

34 | 0,0,1,1,3,4,5,6,6,6,7,7

35 | 0,0,1,1,2,4,5,5,5,5,6,6,

36 | 1,3,4,4,4,5,6,7,7,

37 | 0,1,4,4,6,8

b. Calculate the median and quartiles of these data.

Number of terms = 64

1) Median = 1/2(n + 1)th value

n = 64

= 1/2(64 + 1)th

= 1/2(65)th

= 32.5 th value

This means it is between the 32nd and 33rd value

32nd = 34.5

33rd = 34.6

= 34.5 + 34.6/2

= 69.1/2

= 34.55

2) First Quartile

1/4(n + 1)th value

n = 64

= 1/4(64 + 1)th

= 1/4(65)th

= 16.5th value

This means it is between the 16th and 17th value

16th value = 33.0

17tj value = 33.2

= 33.0 + 33.2/2

= 66.2/2

= 34.55

Q1 --> 33.1

3)Second Quartile = Median

1/2(n + 1)th value

n = 64

= 1/2(64 + 1)th

= 1/2(65)th

= 32.5 th value

This means it is between the 32nd and 33rd value

32nd = 34.5

33rd = 34.6

= 34.5 + 34.6/2

= 69.1/2

= 34.55

Q2 --> 34.55

Third Quartile

3/4(n + 1)th value

n = 64

= 3/4(64 + 1)th

= 3/4(65)th

= 48.75 th value

This means it is towards the 49th value

32nd = 34.5

Hence,

Q3 --> 35.6

Inter Quartile range

Q3 - Q1

= 35.6 - 33.1

= 2.5

8 0

3 years ago

7. Write an inequality: Twenty-two more than four times a number is less

alexandr402 [8]

Answer is C because it's less than.

8 0

3 years ago

Read 2 more answers

16 of 3

Nataliya [291]

107 m

Step-by-step explanation:

Step 1:

Here the microscope magnifies by 10 times.

The initial length is 10.7 m

Step 2:

Let the object be magnified by 10 times . Then, 10 × 10.7 = 107

5 0

3 years ago

Help, please.

scoray [572]

Answer:

d

Step-by-step explanation:

if am not wrong I think the answer is d the last one

3 0

2 years ago

The probability that a student has a Visa card (event V) is .73. The probability that a student has a MasterCard (event M) is .1

snow_lady [41]

We assumed in this answer that the question b is, Are the events V and M independent?

Answer:

(a). The probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ . (b). The events V and M are not independent.

Step-by-step explanation:

The key factor to solve these questions is to know that:

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

We already know from the question the following probabilities:

$\\ P(V) = 0.73$

$\\ P(M) = 0.18$

The probability that a student has both cards is 0.03. It means that the events V AND M occur at the same time. So

$\\ P(V \cap M) = 0.03$

The probability that a student has either a Visa card or a MasterCard

We can interpret this probability as $\\ P(V \cup M)$ or the sum of both events; that is, the probability that one event occurs OR the other.

Thus, having all this information, we can conclude that

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

$\\ P(V \cup M) = 0.73 + 0.18 - 0.03$

$\\ P(V \cup M) = 0.88$

Then, the probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ .

Are the events V and M independent?

A way to solve this question is by using the concept of conditional probabilities.

In Probability, two events are independent when we conclude that

$\\ P(A|B) = P(A)$ [1]

The general formula for a conditional probability or the probability that event A given (or assuming) the event B is as follows:

$\\ P(A|B) = \frac{P(A \cap B)}{P(B)}$

If we use the previous formula to find conditional probabilities of event M given event V or vice-versa, we can conclude that

$\\ P(M|V) = \frac{P(M \cap V)}{P(V)}$

$\\ P(M|V) = \frac{0.03}{0.73}$

$\\ P(M|V) \approx 0.041$

If M were independent from V (according to [1]), we have

$\\ P(M|V) = P(M) = 0.18$

Which is different from we obtained previously;

That is,

$\\ P(M|V) \approx 0.041$

So, the events V and M are not independent.

We can conclude the same if we calculate the probability

$\\ P(V|M)$ , as follows:

$\\ P(V|M) = \frac{P(V \cap M)}{P(M)}$

$\\ P(V|M) = \frac{0.03}{0.18}$

$\\ P(V|M) = 0.1666.....\approx 0.17$

Which is different from

$\\ P(V|M) = P(V) = 0.73$

In the case that both events were independent.

Notice that

$\\ P(V|M)*P(M) = P(M|V)*P(V) = P(V \cap M) = P(M \cap V)$

$\\ \frac{0.03}{0.18}*0.18 = \frac{0.03}{0.73}*0.73 = 0.03 = 0.03$

$\\ 0.03 = 0.03 = 0.03 = 0.03$

3 0

3 years ago