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Ket [755]
3 years ago
13

A jockey will ride horses in 10 different races at a race track today. The number of races that the jockey will win is what type

of random variable
Mathematics
1 answer:
Rasek [7]3 years ago
5 0

Answer:

X is a discrete Random variable where X is the number of races Jockey can win.

Step-by-step explanation:

To find - A jockey will ride horses in 10 different races at a race track today.

The number of races that the jockey will win is what type of random variable ?

Proof -

Given that,

A jockey will ride horses in 10 different races at a race track today.

Let us assume that,

X is the number of races Jockey can win

So,

X can take value 0, 1, 2, 3, ......10

As,

It can take countable finite values

So,

X is a discrete Random variable

∴ we get

X is a discrete Random variable where X is the number of races Jockey can win.

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Is it true that 90% of people will marry their 7th grade love?
Aloiza [94]

Answer:

No

Step-by-step explanation:

No because in forgein countries there is arranged marriages and there are more countries that do that more than the countries that have freedom to marry whoever they want.

3 0
3 years ago
The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 1100lbs.
gizmo_the_mogwai [7]

Answer:

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

<em> P(920≤ x≤1730) = 0.7078 </em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given mean of the Population = 1100 lbs

Standard deviation of the Population = 300 lbs

Let 'X' be the random variable in Normal distribution

Let x₁ = 920

Z = \frac{x-mean}{S.D} = \frac{920-1100}{300} = - 0.6

Let x₂ = 1730

Z = \frac{x-mean}{S.D} = \frac{1730-1100}{300} = 2.1

<u><em>Step(ii)</em></u>

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(x₁≤ x≤x₂) = P(Z₁≤ Z≤ Z₂)

                  = P(-0.6 ≤Z≤2.1)

                  = P(Z≤2.1) - P(Z≤-0.6)

                 = 0.5 + A(2.1) - (0.5 - A(-0.6)

                 =  A(2.1) +A(0.6)               (∵A(-0.6) = A(0.6)

                 =  0.4821 + 0.2257

                 = 0.7078

<u><em>Conclusion:-</em></u>

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

           <em> P(920≤ x≤1730) = 0.7078 </em>

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