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3 years ago

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Mathematics

1 answer:

nasty-shy [4]3 years ago

7 0

Answer:

See Below.

Step-by-step explanation:

Remember that:

If c² > a² + b², then we have an obtuse triangle.
If c² < a² + b², then we have an acute triangle.
And if c² = a² + b², then we have a right triangle.

Where c is the longest side, and a and b are the two remaining sides.

$4^2\text{ is }\textbf{ less than (or $$

$\text{Therefore, }\Delta JKL \text{ is }\textbf{an acute triangle.}$

(16 is less than 9 + 9 or 18.)

$5^2\text{ is } \textbf{equal to (or $=$) } 3^2+4^2$

$\text{Applying the same method, $\Delta ABC$ is }\textbf{ a right triangle.}$

(25 is equal to 9 + 16 = 25.)

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(a) The probability distribution of the random variable X is Binomial.

(b) The mean and standard deviation of the 25-question test are 5 and 2 respectively.

Step-by-step explanation:

The random variable X is defined as the number of correct answers given by a person who is guessing each answer on a 25-question exam.

There are five possible answer for every question.

This implies that the probability of getting a correct answer is:

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Every answer is independent of the others.

(a)

The random variable X has finite number of independent trials (i.e. 25 questions). There are only two outcomes for each trial, i.e. Success = correct answer and Failure = wrong answer. Each trial has the same probability of success (, i.e. P (X) = 0.25).

Thus, the probability distribution of the random variable X is Binomial with parameters n = 25 and p = 0.20.

(b)

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$E(X)=np\\=25\times 0.20\\=5$

Compute the standard deviation of the random variable X as follows:

$SD(X)=\sqrt{np(1-p)}\\=\sqrt{25\times 0.20\times (1-0.20)}\\=2$

Thus, the mean and standard deviation of the 25-question test are 5 and 2 respectively.

(c)

The sample is large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 5

2. n(1 - p) ≥ 5

Check the conditions as follows:

$np=25\times 0.20=5=5\\\\n(1-p)=25\times (1-0.20)=20>5$

Thus, a Normal approximation to binomial can be applied.

So, $X\sim N(5, 2^{2})$

It is provided that the minimum passing score foe the test is such that only 1% of students who are guessing will pass the test.

That is P (X < x) = 0.01.

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The value of z is -2.33.

Compute the value of x as follows:

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Thus, the value of x is 0.34.

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