For a continuous random variable X, P(23 ≤ X ≤ 66) = 0.25 and P(X > 66) = 0.14. Calculate the following probabilities. (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 2 decimal places.)

Answers

Answer 1

Complete question :

For a continuous random variable X, P(23 ≤ X ≤ 66) = 0.25 and P(X > 66) = 0.14. Calculate the following probabilities. (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 2 decimal places.)

a. P(X < 66)

b. P(X < 23)

c. P(X = 66)

Answer:

P(X < 66) = 0.86

P(X < 23) = 0.61

P(X = 66) = 0

Step-by-step explanation:

Given that:

X, P(23 ≤ X ≤ 66) = 0.25 and

P(X > 66) = 0.14 ;

Recall:

P(x ≤ a) + p(x ≥ a) = 1

Then ; P(X < 0.66) = 1 - 0.14 = 0.86

Hence,

P(X < 0.66) = 0.86

P(23 ≤ X ≤ 66) = 0.25

P(X < 66) = 0.86

Hence,

P(X < 23) = 0.86 - 0.25 = 0.61

Probability that a continuous random variable is exactly equal to a particular number equals 0

Hence,

P(X = 66) = 0


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Answer:

12.9

Step-by-step explanation:

A square has even lengths on each side. Since there is 4 sides on a square. We divide perimeter by 4.

51.6/4 = 12.9 inches

The length of each side of the square is 12.9 inches if perimeter of square is 51.6 inches.

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A quadrilateral is defined as a two-dimensional shape with four sides, four vertices, and four angles.

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A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 12 nursing students from Group 1 resulted in a mean score of 59.7 with a standard deviation of 2.8. A random sample of 15 nursing students from Group 2 resulted in a mean score of 64.7 with a standard deviation of 8.3. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let µ1 represent the mean score for Group 1 and µ2 represent the mean score for Group 2. Use a significance level of α = 0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed.Step 1: State the null and alternative hypotheses for the test.Step 2: Compute the value of the t test statistic. Round your answer to three decimal placesStep 3: Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places. Reject or fail to reject your hypothesis?

Answers

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Group 1:

μ1 = 59.7

s1 = 2.8

n1 = sample size = 12

Group 2:

μ2 = 64.7

s2 = 8.3

n2 = sample size = 15

α = 0.1

Assume normal distribution and equ sample variance

A.)

Null and alternative hypothesis

Null : μ1 = μ2

Alternative : μ1 < μ2

B.)

USing the t test

Test statistic :

t = (m1 - m2) / S(√1/n1 + 1/n2)

S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))

S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))

S = 6.4829005

t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)

t = - 5 / 2.5108165

tstat = −1.991384

Decision rule :

If tstat < - tα, (n1+n2-2) ; reject the Null

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From t table :

-t0.1, 25 = - 1.3163

tstat = - 1.9913

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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I hope this helps!

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Step-by-step explanation:

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∠ 53°

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[tex]Hope\\This\\Helps[/tex]

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