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SOVA2 [1]
3 years ago
15

Amanda surveyed 13 students in her class about their heights in inches. Her data are listed below: 58, 59, 55, 52, 57, 59, 62, 5

8, 55, 56, 59, 65, 53
What is the IQR of this data set?
What is the Median (Q2) of this data set?
Mathematics
2 answers:
Anton [14]3 years ago
6 0
Iqr is 4, median is 58 hope that answers your question.
Aleonysh [2.5K]3 years ago
3 0

Answer:

IQR is 4, median is 58.

Step-by-step explanation:

Enter these values in a list, hit stat > calc > 1-var stats. That gives you the 5 number summary. To get IQR, subtract Q3 from Q1.

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Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati
Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly​ selected,  the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c.   D. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X \sim N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})

P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})

P(X < 76) = P(Z< \dfrac{3}{12.5})

P(X < 76) = P(Z< 0.24)

From the standard normal distribution tables,

P(X < 76) = 0.5948

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b.  If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})

P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})

P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})

P( \overline X < 76) = P(Z< 1.2)

From the standard normal distribution tables,

P(\overline X < 76) = 0.8849

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

In order to determine the probability in part (b);  the  normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0
2 years ago
2ln(5x)=8 solve for x
White raven [17]
lnx=log_ex\\----------\\log_ab=c\iff a^c=b\\----------\\Domain:\\5x > 0\ \ \ |divide\ both\ sides\ by\ 5\\x > 0\\D:x\in\mathbb{R^+}\\----------------------\\2ln(5x)=8\ \ \ \ |divide\ both\ sides\ by\ 2\\\\ln(5x)=4\iff e^4=5x\\\\5x=e^4\ \ \ \ \ |divide\ both\ sides\ by\ 5\\\\\boxed{x=\frac{e^4}{5}\in D}
8 0
3 years ago
Read 2 more answers
The coordinates of AABC are A(0,0), B(7,0), and C(1,5). Triangle A'B'C'is the image of AABC rotated counterclockwise about the o
prisoha [69]

Answer:

i dont know

Step-by-step explanation:

that is too hard

3 0
2 years ago
Which relationships have the same constant of proportionality between y and x as the equation y
sdas [7]

Answer:

I'm not too good at this but I got 10/4??

Step-by-step explanation:

please let me know if I'm right!

6 0
3 years ago
Hep me find the measure of angle y
sergiy2304 [10]

Answer:

130

Step-by-step explanation:

first find angle C, which is 180-90-40=50

angle Y=180-50=130

6 0
2 years ago
Read 2 more answers
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