91Ó°ÊÓ

At the point when we're uncertain about the result of an occasion, we can discuss the probabilities of specific results and how likely they are to occur. The number zero indicates the uncertainty of the event and one indicates the certainity.

Given,

The standard deviation $\mathrm{σ}$is $14$minutes.

Sample size n = $49$

Average time (mean)${\mathrm{μ}}_x=145$minutes

The average of $49$races is $\stackrel{-}{X}$

Using the formula and substituting the values,

$\stackrel{¯}{X}~N\left({\mathrm{μ}}_x,\frac{{\mathrm{σ}}_x}{\sqrt{n}}\right)$

$\stackrel{¯}{X}~N\left(145,\frac{14}{\sqrt{49}}\right)$

Hencerole="math" localid="1652282570889" $\stackrel{-}{X}~N(145,2)$

We know,

$\stackrel{-}{X}~N(145,2)$

where$\stackrel{-}{X}$is the average of$49$races

Finding the probability that the runner will average between $142$and$146$minutes in these $49$marathons is,

Using the calculator,

role="math" localid="1652282899504" $$\begin{gathered} P(142<\stackrel{¯}{x}<146)=\text{normalcdf}(142,146,145,2) \\ =0.624 \end{gathered}$$

We know,

$\stackrel{-}{X}~N(145,2)$

where $\stackrel{-}{X}$is the average of $49$races.

For the average of these$49$marathons the $80th$percentile is,

$P(\stackrel{¯}{x}<k)=0.80$

Using the calculator to find k,

$\text{invnorm}(0.80,145,2)=146.68$

Hence the $80th$percentile is$146.68$

Given,

$\stackrel{-}{X}~N(145,2)$

where $\stackrel{-}{X}$is the average of $49$races

We know the median is = the mean of the random variable in a normal distribution

Hence, median = mean = $145$

The median of the average running times is$145$min