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3 years ago

5

If X is a normal random variable with parameters µ = 10 and σ 2 = 36, compute

1 answer:

alex41 [277]3 years ago

6 0

Answer:

Step-by-step explanation:

Given that X is a normal random variable with parameters µ = 10 and σ 2 = 36,

X is N(10, 6)

Or z = $\frac{x-10}{6}$

is N(0,1)

a) P(X > 5),

= $P(Z>-0.8333)\\=0.7977$

(b) P(4 < X < 16),

= $P(|z|$

(c) P(X < 8),

= $P(Z$

(d) P(X < 20),

= $P(Z$

(e) P(X > 16).

=P(Z>-0.6667)

= 0.2524

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3 years ago

Read 2 more answers

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pshichka [43]

Answer:

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed $\frac{dx}{dt}$ = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

$(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=(x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=rx$

$\frac{dh}{dt}=\frac{rx}{h}$

When automobile is 30 miles past the intersection,

For x = 30

$\frac{dh}{dt}=\frac{30r}{h}$

Since $h=\sqrt{d^{2}+(30)^{2}}$

Therefore,

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}$

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

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3 years ago