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4 years ago

Find the probabilities for a standard normal random variable Z.

Mathematics

1 answer:

vampirchik [111]4 years ago

4 0

Answer:

Step-by-step explanation:

find the attachment showing std normal curve symmetrical about y axis.

Equal probabilities on either side of the mean thus the total probability to the right of mean is 0.50

From the table we can find that

a) P(Z>2.5) = 0.5- area lying between 0 and 2.5

= 0.5-0.4938 =0.0062

b) P(1.2<z<2.2) = F(2.2)-F(1.2)

= 0.9861-0.3849

=0.6012

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the annual income of Sam is $37,368 . what is his monthly income if he earns an equal amount every month

max2010maxim [7]

Answer:

$\huge\boxed{\sf \$ 3,114}$

Step-by-step explanation:

Annual Income of Sam = $37,368

To find the monthly income (given that he earns an equal amount of money every month), we can divide the annual income by 12. (Each year has 12 months)

<u>Monthly Income:</u>

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2 years ago

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Step-by-step explanation:

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3 years ago

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Kitty [74]

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3 years ago

Help please ! solve for x and round the answer to the nearest tenth. need it asap , thank you

Jlenok [28]

Answer:

5.2

Step-by-step explanation:

$\frac{17}{13} = \frac{x}{4}$

To solve for x, we want to isolate x. To do this, we need to multiply both sides by 4.

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3 years ago

Read 2 more answers

7. Sky Castle Company is conducting research on 3 independent communities of 12 members each to see how they have been consuming

tresset_1 [31]

Answer:

1.98% probability that at least 11 members are categorized as low risk participants in two of three communities

Step-by-step explanation:

To solve this question, we need to use two separate binomial trials.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of at least 11 members being categorized as low risk participants.

12 members in the sample, so $n = 12$

30% chance to be categorized as high risk participants. So 100-30 = 70% probability of being categorized as low risk participants. So $p = 0.7$

This probability is

$P(X \leq 11) = P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{12,11}.(0.7)^{11}.(0.3)^{1} = 0.0712$

$P(X = 2) = C_{12,12}.(0.7)^{12}.(0.3)^{0} = 0.0138$

$P(X \leq 11) = P(X = 11) + P(X = 12) = 0.0712 + 0.0138 = 0.0850$

What is the probability that at least 11 members are categorized as low risk participants in two of three communities?

For each community, 8.50% probability of at least 11 members being categorized as low risk. So $p = 0.085$

Three comunities, so $n = 3$

This probability is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{3,2}.(0.085)^{2}.(0.915)^{1} = 0.0198$

1.98% probability that at least 11 members are categorized as low risk participants in two of three communities

6 0

3 years ago