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erastova [34]
4 years ago
13

Simplify:

Mathematics
1 answer:
Anna11 [10]4 years ago
4 0

Answer:

1)10y

2)-17z+28

3)29k+42

4)9p-3

5)3s+2

6)w-4

7)-17t+80

8)7t-8

9)2n-8

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Which of the following probabilities is equal to approximately 0. 2957? Use the portion of the standard normal table below to he
Travka [436]

Probability of an event is the measure of its chance of occurrence. The event out of the listed events whose probability is 0.2957 is given by : Option C: P(0.25 \leq Z \leq 1.25)

<h3>How to get the z scores?</h3>

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z-score.

If we have

X \sim N(\mu, \sigma)

(X is following normal distribution with mean \mu and standard deviation \sigma )

then it can be converted to standard normal distribution as

Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

P(Z \leq z) = P(Z < z) )

Also, know that if we look for Z = z in z tables, the p-value we get is

P(Z \leq z) = \rm p \: value

Using the z-table, we get the needed probabilities as:

Case 1:

P(-1.25 \leq Z \leq 0.25) = P(Z \leq 0.25) - P(Z \leq -1.25) \approx 0.5987 - 0.1056 = 0.4931

Case 2:

P(-1.25 \leq Z \leq 0.75) = P(Z \leq 0.75) - P(Z \leq -1.25) \approx 0.7734- 0.1056=0.6678

Case 3:

P(0.25 \leq Z \leq 1.25) = P(Z \leq 1.25) - P(Z \leq 0.25) \approx 0.8944 - 0.5987=0.2957

Case 4:

P(0.75 \leq Z \leq 1.25) = P(Z \leq 1.25) - P(Z \leq 0.75) \approx 0.8944 - 0.7734 =0.1210

Thus, the event out of the listed events whose probability is 0.2957 is given by : Option C: P(0.25 \leq Z \leq 1.25)

Learn more about z-scores here:

brainly.com/question/13299273

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