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oksano4ka [1.4K]

3 years ago

14

According to the empirical rule, if the data form a bell-shaped normal distribution, __________ of the observations will fall wi

thin two standard deviations around the mean.

1 answer:

Kaylis [27]3 years ago

5 0

Answer:

95%

Step-by-step explanation:

The empirical rule states that if data follows normal distribution then the percentage of observations falls within one, two and three standard deviation around the mean are

i) 68% falls within one standard deviation

ii) 95% falls within two standard deviation

iii) 99.7% falls within three standard deviation.

Hence 95% of the observations will fall within two standard deviations around the mean if the data follows normal distribution.

You might be interested in

What is 26/48 simplest form

Phoenix [80]

Answer:

13/14

Step-by-step explanation:

The GCF between 26 and 48 is 2. So divide 26/2=12 and 48/2=13 13/14. And after that you can not simplify anymore so 13/14 is 26/48 in it's simplest form.

Hope it helps!

5 0

3 years ago

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

2 years ago

Solve a,b,c,d and e.

Westkost [7]

Answer:

Step-by-step explanation:

a)

428721

Place of 2's 10s and 10,000s

Therefore its value is 20 and 20,000

Product of the place value = 20 x 20,000 = 4,00,000

b)

37,20,861

Place of 7 is 1,00,000

Therefore the place value is 7,00,000

c)

Greatest 7 digit number is 99,99,999

Adding 1 to it = 99,99,999 + 1 = 1,00,00,000

d)

85642 = 80000 + 5000 + 600 + 40 +2

e)

round off 85642 to nearest thousand = 86,000

7 0

2 years ago

HELPP!!!! 100 points. Which expression is equivalent

devlian [24]

Answer:

$\Large \boxed{\sf A}$

Step-by-step explanation:

$\displaystyle 2x\sqrt{44x}-2\sqrt{11x^3}$

Let<em> x</em> = 2 because x > 0

$\displaystyle 2(2)\sqrt{44(2)}-2\sqrt{11(2)^3}=4\sqrt{22}$

See which expression is equal to 4√22 when <em>x</em> = 2

$2(2)\sqrt{11(2)}=4\sqrt{22}$

The expression is option A

8 0

2 years ago

Read 2 more answers

Use the integral test to determine whether the series is convergent or divergent. [infinity] n = 2 n2 n3 + 1 Evaluate the follow

xxMikexx [17]

I think the given series is

$\displaystyle\sum_{n=2}^\infty \frac{n^2}{n^3+1}$

You can use the integral test because the summand is clearly positive and decreasing. Then

$\displaystyle\sum_{n=2}^\infty\frac{n^2}{n^3+1} > \int_2^\infty\frac{x^2}{x^3+1}\,\mathrm dx$

Substitute <em>u</em> = <em>x</em> ³ + 1 and d<em>u</em> = 3<em>x</em> ² d<em>x</em>, so the integral becomes

$\displaystyle \int_2^\infty\frac{x^2}{x^3+1}\,\mathrm dx = \frac13\int_9^\infty\frac{\mathrm du}u = \frac13\ln(u)\bigg|_{u=9}^{u\to\infty}$

As <em>u</em> approaches infinity, we have ln(<em>u</em>) also approaching infinity (whereas 1/3 ln(9) is finite), so the integral and hence the sum diverges.

5 0

2 years ago