Chapter 11.1, Problem 24E

To determine

To find: the probability that at most $50$ seats were observed.

Answer to Problem 24E

The result is $P\left(z\le -1.6\right)=0.0548$

Explanation of Solution

Given:

Mean is $73$ seals

Standard deviation is $14.1$ scals.

Calculation:

Given that the number of seals observed are normally distributed with a mean of $73$ seals and standard deviation of $14.1$ seals. That is $\mu =73$ and $\sigma =14.1$ .

find the probability that at most $50$ seals were observed during a randomly chosen survey.

So, find the $z-score$ corresponding to an $x-value$ of $50$ .

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{50-73}{14.1}\\ =-1.6\end{array}$

Thus, required probability is given by the value of $P\left(z\le -1.6\right)$

Now, use the standard normal table to find $P\left(z\le -1.6\right)$ .

To find $P\left(z\le -1.6\right)$ using standard normal table, find the value where row $-1$ and column $.6$ intersects that is $.0548$ . This value is the probability $P\left(z\le -1.6\right)$ .

Standard Normal Table
$z$	$.0$	$.1$	$.2$	$.3$	$.4$	$.5$	$.6$	$.7$	$.8$	$.9$
$-3$	$.0013$	$.0010$	$.0007$	$.0005$	$.0003$	$.0002$	$.0002$	$.0001$	$.0001$	$.0000+$
$-2$	$.0228$	$.0179$	$.0139$	$.0107$	$.0082$	$.0062$	$.0047$	$.0035$	$.0026$	$.0019$
$-1$	$.1587$	$.1357$	$.1151$	$.0968$	$.0808$	$.0668$	$.0548$	$.0446$	$.0359$	$.0287$
$-0$	$.5000$	$.4602$	$.4207$	$.3821$	$.3446$	$.3085$	$.2743$	$.2420$	$.2119$	$.1841$
$0$	$.5000$	$.5398$	$.5793$	$.6179$	$.6554$	$.6915$	$.7257$	$.7580$	$.7881$	$.8159$
$1$	$.8413$	$.8643$	$.8849$	$.9032$	$.9192$	$.9332$	$.9452$	$.9554$	$.9641$	$.9713$
$2$	$.9772$	$.9821$	$.9861$	$.9893$	$.9918$	$.9938$	$.9953$	$.9965$	$.9974$	$.9981$
$3$	$.9987$	$.9990$	$.9993$	$.9995$	$.9997$	$.9998$	$.9998$	$.9999$	$.9999$	$1.000-$

The probability that at most $50$ seals were observed during survey is given by

  $P\left(z\le -1.6\right)=0.0548$

Conclusion:

Hence, the result is $P\left(z\le -1.6\right)=0.0548$