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Suppose that a random variableXhas the binomial distribution with parametersn=15 andp=0.5. Find Pr(X <6).

Short Answer

Expert verified

The probability of X being less than 6 is 0.1509

Step by step solution

01

Given information

The random variable X follows the binomial distribution with parameters \(n = 15\) and \(p = 0.5\)

02

Compute the probability

The probability function of the binomially distributed random variable X is given as,

\(f\left( x \right) = \left\{ \begin{array}{l}\left( \begin{array}{l}n\\x\end{array} \right){p^x}{\left( {1 - p} \right)^{n - x}};\;x = 0,1,...,n\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;;\;otherwise\end{array} \right.\)

where there are n trials with p probability of success in each trial.

The required probability is computed as,

\(\begin{aligned}{c}P\left( {X < 6} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) + ... + P\left( {X = 5} \right)\\ &= \left( {\left( \begin{aligned}{l}15\\0\end{aligned} \right){{\left( {0.5} \right)}^0}{{\left( {1 - 0.5} \right)}^{15 - 0}}} \right) + \left( {\left( \begin{aligned}{l}15\\1\end{aligned} \right){{\left( {0.5} \right)}^1}{{\left( {1 - 0.5} \right)}^{15 - 1}}} \right) + ... + \left( {\left( \begin{aligned}{l}15\\5\end{aligned} \right){{\left( {0.5} \right)}^5}{{\left( {1 - 0.5} \right)}^{15 - 5}}} \right)\\ &= 0.00003 + 0.00046 + ... + 0.09164\\& = 0.1509\end{aligned}\)

Therefore, the required probability is approximately 0.1509.

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Most popular questions from this chapter

A civil engineer is studying a left-turn lane that is long enough to hold seven cars. LetXbe the number of cars in the lane at the end of a randomly chosen red light. The engineer believes that the probability thatX=xis proportional to(x+1)(8−x)forx=0, . . . ,7 (the possible values ofX).

a. Find the p.f. ofX.

b. Find the probability thatXwill be at least 5.

Question: Suppose that the joint p.d.f. of two random variablesXandYis as follows:

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}c\left( {{x^2} + y} \right)\,\,\,\,for\,0 \le y \le 1 - {x^2}\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\)

Determine (a) the value of the constantc;

\(\begin{array}{l}\left( {\bf{b}} \right)\,{\bf{Pr}}\left( {{\bf{0}} \le {\bf{X}} \le {\bf{1/2}}} \right){\bf{;}}\,\left( {\bf{c}} \right)\,{\bf{Pr}}\left( {{\bf{Y}} \le {\bf{X + 1}}} \right)\\\left( {\bf{d}} \right)\,{\bf{Pr}}\left( {{\bf{Y = }}{{\bf{X}}^{\bf{2}}}} \right)\end{array}\)

Suppose that thenrandom variablesX1, . . . , Xnform a random sample from a continuous distribution for which the p.d.f. isf. Determine the probability that at leastk of thesenrandom variables will lie in a specified intervala≤x≤b.

Let Xbe a random variable with the p.d.f. specified in Example 3.2.6. Compute Pr(X≤8/27).

Question:Suppose that in a certain drug the concentration of aparticular chemical is a random variable with a continuousdistribution for which the p.d.f.gis as follows:

\({\bf{g}}\left( {\bf{x}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{\bf{3}}}{{\bf{8}}}{{\bf{x}}^{\bf{2}}}\;{\bf{for}}\;{\bf{0}} \le {\bf{x}} \le {\bf{2}}\\{\bf{0}}\;{\bf{otherwise}}\end{array} \right.\)

Suppose that the concentrationsXandYof the chemicalin two separate batches of the drug are independent randomvariables for each of which the p.d.f. isg. Determine

(a) the joint p.d.f.of X andY;

(b) Pr(X=Y);

(c) Pr(X >Y );

(d) Pr(X+Y≤1).

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