91影视

The uniform distribution on the interval $1鈮�x鈮�5$ has been used to model the distribution.

Such that

$a=1$

And

$b=5$

A density curve is always on or above the horizontal axis.

The density curve is reciprocal to the difference in the borders in a uniform distribution.

On the interval between the boundaries,

$f\left(x\right)=\frac{1}{b-a}=\frac{1}{5-1}=\frac{1}{4}=0.25$

With

$1鈮�x鈮�5$

$f\left(x\right)$ represents the height of the density curve.

Thus, the height of the density curve is $0.25$

The likelihood that the duration until light flashes is between $3.75<X<5$ equals the area beneath the density curve between $3.75<X<5$ (maximum time is $5$ seconds).

Note that

The rectangle will be the area beneath the density curve.

With

Width, $W=5鈭�3.75=1.25$

And

Height, $H=f\left(x\right)=0.25$

Then

Therefore,

$31.25\%$ of the time that the light flashes more than $3.75$ seconds after the subject clicks 鈥淪tart鈥�.

$38$ percent of the data values should be smaller than the $38th$ percentile, according to the attribute for the $38th$ percentile.

Let

$x$be the $38th$percentile.

The likelihood that the time is between the lower boundary and $x$ will be the area beneath the density curve between $1$ and $x$

Note that

The area underneath the density curve will be the rectangle.

With

Width, $W=x鈭�1$

And

Height, $H=f\left(x\right)=0.25$

Then

We know that

$x$ is the $70th$ percentile.

Then

The probability has to be equal to $38\%$ or $0.38$

$0.25x鈭�0.25=0.38$

Add $0.25$ to both sides.

$0.25x=0.63$

Divide the above equation by $0.1$

That becomes

$x=\frac{0.63}{0.25}=2.52$

Therefore,

The $38th$ percentile of the distribution will be $2.52$ seconds, implying that the participants must click the button in fewer than $2.52$ seconds 38 percent of the time.