Consider the density curve plotted below:1920212223240.0250.050.0750.10.1250.150.1750.20.2250.250.275XPDF(X)Density CurveFind P(X≤22) : 0.1Find P(X>21) : Calculate the following. Q1: median: Q3: IQR:

Consider the density curve plotted below1920212223240025005007501012501501750202250250275XPDFXDensity CurveFind PX22 01Find PXgt21 Calculate the following Q1 me class=

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Explanation

In this problem, we have a graph of the PDF (Probability Density Function). To compute probabilities in a certain interval (a, b), we must integrate this function from x = a to x = b.

(1) P(X ≤ 22)

We integrate the function from x = -∞ to x = 22, we get:

[tex]\begin{gathered} P(X\text{ }≤\text{ }22)=\int_{-\infty}^{22}dx\cdot PDF(x) \\ =\int_{-\infty}^{20}dx\cdot PDF(x)+\int_{20}^{22}dx\cdot PDF(x) \\ =\int_{-\infty}^{20}dx\cdot0+\int_{20}^{22}dx\cdot0.25 \\ =0+0.25\cdot(22-20) \\ =0.25\cdot2 \\ =0.5. \end{gathered}[/tex]

We separated the integral to use the data from the graph.

(2) P(X > 21)

We integrate the function from x = 21 to x = ∞, we get:

[tex]\begin{gathered} P(X>21)=\int_{21}^{\infty}dx\cdot PDF(x) \\ =\int_{21}^{24}dx\cdot PDF(x)+\int_{24}^{\infty}dx\cdot PDF(x) \\ =\int_{21}^{24}dx\cdot0.25+\int_{24}^{\infty}dx\cdot0 \\ =0.25\cdot(24-21)+0 \\ =0.25\cdot3 \\ =0.75. \end{gathered}[/tex]

(3) The Q1 is the value x = a of the interval (-∞, a) that gives a probability equal to 0.25. So we must find x such that:

[tex]P(XUsing the data of the graph, we have:[tex]\begin{gathered} \int_{-\infty}^adx\cdot PDF(x)+\int_{20}^adx\cdot PDF(x)=0.25, \\ \int_{-\infty}^{20}dx\cdot0+\int_{20}^adx\cdot0.25=0.25, \\ 0.25\cdot(a-20)=0.25, \\ a-20=\frac{0.25}{0.25}, \\ a-20=1, \\ a=21. \end{gathered}[/tex]

(4) The median is the value x = a of the interval (-∞, a) that gives a probability equal to 0.5. Proceeding as before, we have:

[tex]\begin{gathered} \int_{-\infty}^adx\cdot PDF(x)+\int_{20}^adx\cdot PDF(x)=0.5, \\ \int_{-\infty}^{20}dx\cdot0+\int_{20}^adx\cdot0.25=0.5, \\ 0.25\cdot(a-20)=0.5, \\ a-20=\frac{0.5}{0.25}, \\ a-20=2, \\ a=22. \end{gathered}[/tex]

(5) The Q3 is the value x = a of the interval (-∞, a) that gives a probability equal to 0.75. Proceeding as before, we have:

[tex]\begin{gathered} \int_{-\infty}^adx\cdot PDF(x)+\int_{20}^adx\cdot PDF(x)=0.75, \\ \int_{-\infty}^{20}dx\cdot0+\int_{20}^adx\cdot0.25=0.75, \\ 0.25\cdot(a-20)=0.75, \\ a-20=\frac{0.75}{0.25}, \\ a-20=3, \\ a=23. \end{gathered}[/tex]

(6) The IQR is given by the difference between Q3 and Q1. Using the results from above, we get:

[tex]IQR=Q3-Q1=23-21=2.[/tex]Answer

• P(X ≤ 22) = 0.5

,

• P(X > 21) = 0.75

,

• Q1 = 21

,

• median = 22

,

• Q3 = 23

,

• IQR = 2

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