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vazorg [7]
3 years ago
12

Given that x is a normally distributed random variable with a mean of 50 and a standard deviation of 2, find the probability tha

t x is between 47 and 54.
Mathematics
1 answer:
Shkiper50 [21]3 years ago
5 0

Let X be the normal variable with mean μ =50 and standard deviation σ =2

We have to find probability that x is between 47 and 54

P(47 < X < 54) = P(X < 54) - P(X < 47)

= P(\frac{x-mean}{standard deviation} - P(\frac{x-mean}{standard deviation}

= P(Z < 2) - P(Z < -1.5)

Using standard normal z score table to find probabilities we get

P(Z < 2) = 0.9772

P(Z < -1.5) = 0.0668

P(47 < X < 54) = P(Z < 2) - P(Z < -1.5)

= 0.9772 - 0.0668

P(47 < X < 54) = 0.9104

The probability that x is between 47 and 54 is 0.9104

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PLEASE HELP NEED THIS DONE ASAP
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Answer:

The area of rhombus PQRS is 120 m.

Step-by-step explanation:

Consider the rhombus PQRS.

All the sides of a rhombus are equal.

Hence, PQ = QR = RS = SP = 13 m

The diagonals PR and QS bisect each other.

Let the point at of intersection of the two diagonals be denoted by <em>X</em>.

Consider the triangle QXR.

QR = 13 m

XR = 12 m

The triangle QXR is a right angled triangle.

Using the Pythagorean theorem compute the length of QX as follows:

QR² = XR² + QX²

QX² = QR² - XR²

       = 13² - 12²

       = 25

 QX = √25

       = 5 m

The measure of the two diagonals are:

PR = 2 × XR = 2 × 12 = 24 m

QS = 2 × QX = 2 × 5 = 10 m

The area of a rhombus is:

\text{Area}=\frac{1}{2}\times d_{1}\times d_{2}

Compute the area of rhombus PQRS as follows:

\text{Area}=\frac{1}{2}\times PR\times QS

        =\frac{1}{2}\times 24\times 10\\\\=120

Thus, the area of rhombus PQRS is 120 m.

5 0
3 years ago
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