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Sergeu [11.5K]
4 years ago
9

A set of data has a normal distribution with a mean of 29 and a standard deviation of 4. Find the percent of data within each in

terval.
27. from 25 to 33

a. 42%
b. 68%
c. 13%
d. 88%



28. from 21 to 25

a. 28.5%
b. 54.5%
c. 13.5%
d. 15.5%


29. greater than 29

a. 50%
b. 75%
c. 25%
d. 100%



30. less than 21

a. 6.5%
b. 10.5%
c. 1.5%
d. 2.5%
Mathematics
1 answer:
Nikolay [14]4 years ago
7 0
Let x be the population distribution.
p(25 ≤ x ≤ 33) = p((25 - 29)/4 ≤ z ≤ (33 - 29)/4) = p(-1 ≤ z ≤ 1) = p(z ≤ 1) - p(z ≤ -1) = p(z ≤ 1) - [1 - p(z ≤ 1)] = 2p(z ≤ 1) - 1 = 2(0.84134) - 1 = 0.68268 = 68%

p(21 ≤ x ≤ 25) = p((21 - 29)/4 ≤ z ≤ (25 - 29)/4) = p(-2 ≤ z ≤ -1) = p(z ≤ -1) - p(z ≤ -2) = [1 - p(z ≤ 1)] - [1 - p(z ≤ 2)] = p(z ≤ 2) - p(z ≤ 1) = 0.97725 - 0.84134 = 0.13591 = 13.5%


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The volume of a shampoo filled into a container is uniformly distributed between 370 and 385 milliliters. (a) What are the mean
Olenka [21]

Answer:

(a) mean = 377.5ml. standard deviation = 4.33

(b) 0.333

(c) 384.25 ml

Step-by-step explanation:

(a)The mean:

\mu = \frac{385 + 370}{2} = 377.5 ml

The standard deviation:

\sigma = \frac{385 - 370}{\sqrt{12}} = 4.33

(b)

P(x < 375) = \frac{375 - 370}{385 - 370} = \frac{1}{3} \approx 0.333

(c)P(x > m) = 0.95

\frac{m - 370}{385 - 370} = 0.95

m - 370 = 14.25

m = 384.25ml

So the volumn of shampoo should be 384.25ml to exceed 95% of containers.

7 0
3 years ago
Determine whether each of the functions log(n + 1) and log(n2 + 1) is o(log n)
larisa86 [58]
Assuming the order required is as n-> inf.

As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.

We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n)  which is still o(log(n)).

So yes, both are o(log(n)).

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8 0
3 years ago
(-9)4 what is (-9) to the power of for is
Luba_88 [7]

Answer:

6,561

Step-by-step explanation:

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8 0
4 years ago
In one town, the number of burglaries in a week has a poisson distribution with a mean of 1.9. find the probability that in a ra
tester [92]

Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9

We have to find the probability that in a randomly selected week the number of burglaries is at least three.

P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........

= 1 - P(X < 3)

= 1 - [ P(X=2) + P(X=1) + P(X=0)]

The Poisson probability at X=k is given by

P(X=k) = \frac{e^{-mean} mean^{x}}{x!}

Using this formula probability of X=2,1,0 with mean = 1.9 is

P(X=2) = \frac{e^{-1.9} 1.9^{2}}{2!}

P(X=2) = \frac{0.1495 * 3.61}{2}

P(X=2) = 0.2698

P(X=1) = \frac{e^{-1.9} 1.9^{1}}{1!}

P(X=1) = \frac{0.1495 * 1.9}{1}

P(X=1) = 0.2841

P(X=0) = \frac{e^{-1.9} 1.9^{0}}{0!}

P(X=0) = \frac{0.1495 * 1}{1}

P(X=0) = 0.1495

The probability that at least three will become

P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]

= 1 - [0.2698 + 0.2841 + 0.1495]

= 1 - 0.7034

P(X ≥ 3 ) = 0.2966

The probability that in a randomly selected week the number of burglaries is at least three is 0.2966

5 0
3 years ago
A system of equations has infinitely many solutions. if 2y-4x=6 is one of the equations, which could be the other equations?
Romashka [77]
You have to do rearranging formulae to find that out so basically solve for x
8 0
4 years ago
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