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fomenos
3 years ago
9

The number 0.6064 represents the area under the standard normal curve below a particular z-score. What is the Z-score? Can you p

lease show me how this question works and to express the answer as a decimal to the nearest hundredth. Thanks
Mathematics
2 answers:
Rudiy273 years ago
4 0
P(z < x) = 0.6064 - 0.5 = 0.1064

From the normal distribution table
P(z < 0.27) = 0.1064

Therefore the z-score is 0.27


SashulF [63]3 years ago
4 0

Answer:

z=0.27

Step-by-step explanation:

Let x represent the z-score.

We have been given that the The number 0.6064 represents the area under the standard normal curve below a particular z-score.

Since the area under normal distribution curve that corresponds to our z-score is 0.6064, so we can set an equation as:

P(z

From normal distribution table we get,

P(z

Since the area 0.6064 under normal distribution corresponds to z-score of 0.27, therefore, our z-score s 0.27.

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<em>41 degrees</em>

Step-by-step explanation:

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What’s the inverse equation? And how you solved it step by step please!
Korvikt [17]

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The inverse is (x^2-5)/2

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y =sqrt(2x+5)

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x = sqrt(2y+5)

Solve for y

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3 years ago
The output of a chemical process is continually monitored to ensure that the concentration remains within acceptable limits. Whe
marissa [1.9K]

Answer:

a)  0.31 = 31%

b) 0.03 = 3%

c) 0.36 = 36%

d) 2 times

Step-by-step explanation:

If F_X(x) is the cumulative distribution function of the random variable X, then by definition the probability P of the random variable is given by

P(X \leq x) = F_X(x)

If additionally the random variable is discrete (only has non-negative integers as outcomes as is the case in this problem) then

P(X=a)=F_X(a)-\lim_{x \to a^-}F_X(x)

a)

We are looking for P(X<2)

P(X < 2) = P(X\leq 2)-P(X=2)=F_X(2)-\lim_{x \to 2^-}F_X(x)=0.84-0.53=0.31

b)

In this case we want P(X>3)

P(X >3) = 1-P(X\leq 3)=1-F_X(3)=1-0.97=0.03

c)

Now, we are interested in P(X=1)

P(X =1) =F_X(1)-\lim_{x \to 1^-}F_X(x)=0.53-0.17=0.36

d)

The expected number of times that the process is recalibrated during the week is the expected value of the probability distribution:

P(X=1)+2P(X=2)+...+nP(X=n)+...

But it is easy to see that P(X=n) = 0 if n is an integer >4

So, the expected value is

P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)

We already have P(X=1) and P(X=2). Let's compute the rest

P(X =3) =F_X(3)-\lim_{x \to 3^-}F_X(x)=0.97-0.84=0.13

P(X =4) =F_X(4)-\lim_{x \to 4^-}F_X(x)=1-0.97=0.03

and the expected value is

0.36 + 2*0.53+3*0.13+4*0.03= 1.93 = 2 times rounding to the nearest integer.

8 0
4 years ago
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