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dezoksy [38]
3 years ago
9

Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all an

swers to four decimal places.)P(z 2.37)
Mathematics
1 answer:
Lorico [155]3 years ago
4 0

Answer:

P(Z < 2.37) = 0.9911.

Step-by-step explanation:

We are given that Let z denote a random variable that has a standard normal distribution.

Let Z = a random variable

So, Z ~ Standard Normal(0, 1)

As we know that the standard normal distribution has a mean of 0 and variance equal to 1.

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

where, \mu = mean = 0

           \sigma = standard deviation = 1

Now, the probability that z has a value less than 2.37 is given by = P(Z < 2.37)

        P(Z < 2.37) = P(Z < \frac{2.37-0}{1} ) = P(Z < 2.37) = 0.9911

The above probability is calculated by looking at the value of x = 2.37 in the z table which has an area of 0.9911.

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Suppose that from the past experience a professor knows that the test score of a student taking his final examination is a rando
DENIUS [597]

Answer:

n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =73,\sigma =10.5)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

We want to find the value of n that satisfy this condition:

P(71.5 < \bar X

And we can use the z score formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we have this:

P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z

And we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94

And by properties of the normal distribution we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z

If we solve for P(Z we got:

P(Z

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: z=-1.881

And using this we have this equality:

-1.881 = -0.14286 \sqrt{n}

If we solve for \sqrt{n} we got:

\sqrt{n} = \frac{-1.881}{-0.14286}=13.167

And then n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

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3 years ago
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xxMikexx [17]

Answer:

yes

yes

no

Step-by-step explanation:

4 0
4 years ago
T3=-6.5 and t5=-3.9 determine d
nekit [7.7K]
t_3=-6.5;\ t_5=-3.9\\\\t_5-t_3=2d\\\\2d=-3.9-(-6.5)\\\\2d=-3.9+6.5\\\\2d=2.6\ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{d=1.3}
4 0
3 years ago
Regina claims that gold has a mass almost 20 times more than water.
Butoxors [25]

The atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu, also known as daltons, D). The atomic mass is a weighted average of all of the isotopes of that element, in which the mass of each isotope is multiplied by the abundance of that particular isotope. (Atomic mass is also referred to as atomic weight, but the term "mass" is more accurate.)

7 0
3 years ago
HELP ASAP ITS EASY DEGREE CR AP JUST PLS ILL GIVE BRAINIEST NO GUESSING PLSSSSS
lana [24]

Answer:

110°

Step-by-step explanation:

A+B+C=180

48+28+C= 180

70+C=180

subtract 70 form both sides

C= 180

plz mark me brainliest. :0

5 0
3 years ago
Read 2 more answers
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