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3 years ago

Which statements are true about the interquartile range? Select all that apply.

Mathematics

2 answers:

kap26 [50]3 years ago

4 0

Is there a picture? there's not enough info

Natalka [10]3 years ago

4 0

Answer:

no picture!

Step-by-step explanation:

when typing a question, below is a circle with a plus button, this allows to add an article, picture or a document of your work that you need anything help with. hope this helps!

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One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo

almond37 [142]

Answer:

(1) P( $\bar X$ < 26 inches) = 0.0436

(2) P(27.5 inches < $\bar X$ < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let $\bar X$ = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

Z = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean vertical leap = 28 inches

$\sigma$ = standard deviation = 7 inches

n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P( $\bar X$ < 26 inches)

P( $\bar X$ < 26) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{26-28}{\frac{7}{\sqrt{36} } }$ ) = P(Z < -1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < $\bar X$ < 28.5 inches) = P( $\bar X$ < 28.5 inches) - P( $\bar X$ $\leq$ 27.5 inches)

P( $\bar X$ < 28.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{28.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z < 0.36) = 0.64058 {using z table}

P( $\bar X$ $\leq$ 27.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{27.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z $\leq$ -0.36) = 1 - P(Z < 0.36)

= 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < $\bar X$ < 28.5 inches) = 0.64058 - 0.35942 = 0.2812

6 0

3 years ago

HbhhhhvbvbbhHzhzhhshhjzjzbsjzjsjjsujenwja

sineoko [7]

Answer:

hgfhdgsgzfzsgfzgzfffdfdsfdfgfghgf

Step-by-step explanation:

...

6 0

2 years ago

According to postal regulations, a carton is classified as "oversized" if the sum of its height and girth (the perimeter of its

Amanda [17]

Answer:

V(max) = 8712.07 in³

Dimensions:

x (side of the square base) = 16.33 in

girth = 65.32 in

height = 32.67 in

Step-by-step explanation:

Let

x = side of the square base

h = the height of the postal

Then according to problem statement we have:

girth = 4*x (perimeter of the base)

and

4* x + h = 98 (at the most) so h = 98 - 4x (1)

Then

V = x²*h

V = x²* ( 98 - 4x)

V(x) = 98*x² - 4x³

Taking dervatives (both menbers of the equation we have:

V´(x) = 196 x - 12 x² ⇒ V´(x) = 0

196x - 12x² = 0 first root of the equation x = 0

Then 196 -12x = 0 12x = 196 x = 196/12

x = 16,33 in ⇒ girth = 4 * (16.33) ⇒ girth = 65.32 in

and from equation (1)

y = 98 - 4x ⇒ y = 98 -4 (16,33)

y = 32.67 in

and maximun volume of a carton V is

V(max) = (16,33)²* 32,67

V(max) = 8712.07 in³

3 0

3 years ago

ComeShakeYourBodyBabyDoTheCongaCantHoldMyselfAnyLongerComeShakeYourBodyAnyLonger

Helga [31]

Answer:

hello

Step-by-step explanation:

i get points for this so im just gonna answer :3

5 0

2 years ago

Polynomial that represents the area of the square (big ides math)

IrinaVladis [17]

Answer:

49n^2-70n+25

Step-by-step explanation:

(7n-5)(7n-5)

FOIL it: 49n^2-35n-35n+25

combine like terms: 49n^2-70n+25

4 0

3 years ago

Which statements are true about the interquartile range? Select all that apply.​

Which statements are true about the interquartile range? Select all that apply.