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Serhud [2]
3 years ago
6

Does the mean of a normal distribution is always positive? How about the standard deviation?

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

Answer:

Mean of a Normal Distribution:

The mean of normal distribution is not always positive, it is equally distributed around mean,mode and median and can be any value from ranging from negative to positive to infinity.

Standard Deviation:

For the standard deviation, it can not be negative. It can only be equal to any positive values i.e., values \geq 0.

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Marrrta [24]
For the numbers below find the percentile rank<span> (the % of individuals scoring below). ... For the numbers below find percent of cases falling between the two </span>z-scores<span>. ... The patient </span>scores<span> a 45 on the test (mean = </span>52, standard deviation of 5.<span>Find the </span>z-scores<span> corresponding to each of the following values: a) A score ... On one subtest, the </span>raw scores<span> have a mean of 35 and a standard deviation of 6. Assuming ... a) Separate the highest 30% from the rest of the distribution. .</span>52<span> ... For the </span>z-scores<span> below, find the </span>percentile rank<span> (percent of individuals scoring below):.</span>
8 0
3 years ago
True or false??????​
Katyanochek1 [597]

Answer:

true

Step-by-step explanation:

\sqrt{25}  = ±5

6 0
3 years ago
Sense the pic was blurry last time
AfilCa [17]
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6 0
3 years ago
How to solve number 44
solniwko [45]
OK. 

Let's say a member and a non member each visit the garden ' V ' times.

The non-member's cost for each visit is  $6 .
The non member's cost for ' V ' visits is  6 V .
His total cost for the year is  6 V .

The member's cost for each visit is $3.
The member's cost for ' V ' visits is  3V .
His total cost for the year is  3V  +  the  $24 to join.

We want to know what ' V ' is  (how many times each one can visit)
if their total costs are the same.

So let's just write an equation that SAYS their costs are the same,
and see what ' V ' turns out to be.

             Non-member's cost for the year =  Member's cost for the year

                                                            6 V  =  3 V + 24

Subtract  3V  from each side:          3 V  =            24

Divide each side by  3 :                       V  =              8  .

-- If they both visit the garden 1, 2, 3, 4, 5, 6, or 7 times in the year,
the member will spend MORE than the non member.

-- If they both visit the garden 8 times in the year,
they'll both spend the same amount.  ($48)

-- If they both visit the garden MORE than 8 times in the year,
the member will spend LESS than the non-member.
______________________________________________ 

That was the algebra way to do it.

Now here is the cheap, sleazy, logical, easy way to do it:

The non-member spends (6 - 3) = $3 MORE than the member for each visit ?

After how many visits does the $3 more each time add up to the $24 that
it cost the member to join for the year ?

           $24 / $3  =  8 visits  .
5 0
4 years ago
Let X denote the number of bars of service on your cell phone whenever you are at an intersection with the following probabiliti
ZanzabumX [31]

Answer:

a. P(2 \leq X \leq 3) = 0.5

b. P(X \geq 1) = 0.9

c. P(X < 2) = 0.25

d. P(X > 3) = 0.25

Step-by-step explanation:

We are given the following distribution:

P(X = 0) = 0.1

P(X = 1) = 0.15

P(X = 2) = 0.25

P(X = 3) = 0.25

P(X = 4) = 0.15

P(X = 5) = 0.1

a. Two or three bars

P(2 \leq X \leq 3) = P(X = 2) + P(X = 3) = 0.25 + 0.25 = 0.5

Thus:

P(2 \leq X \leq 3) = 0.5

b. At least one bar

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1 = 0.9

Thus:

P(X \geq 1) = 0.9

c. Fewer than two bars

P(X < 2) = P(X = 0) + P(X = 1) = 0.1 + 0.15 = 0.25

Thus:

P(X < 2) = 0.25

d. More than three bars

P(X > 3) = P(X = 4) + P(X = 5) = 0.15 + 0.1 = 0.25

Thus:

P(X > 3) = 0.25

7 0
3 years ago
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