91影视

The standard deviation is a statistic that is calculated as the square root of the variance and measures the dispersion of a dataset relative to its mean. The standard deviation is calculated as the square root of variance by calculating the deviation of each data point from the mean.

Let the birth weights of the babies be represented by an RV X then its mean and standard deviation are given as:

(a) The probability that the birth weight of a randomly selected baby of this type exceeds \(4000\;{\rm{g}}\)can be represented as \(P(4000 < X)\). Standardizing gives:

\(4000 < X\)

if and only if

\(\begin{array}{c}\frac{{4000 - 3432}}{{482}} < \frac{{X - 3432}}{{482}}\\\frac{{568}}{{482}} < \frac{{X - 3432}}{{482}}\\1.18 < Z\end{array}\)

Thus

\(P(4000 < X) = P(1.18 < Z)\)

Here Z is a standard normal distribution rv with cdf \(\phi (z)\). Hence, we can write

\(P(4000 < X) = 1 - \phi (1.18)\)

To get it, \(\phi (1.18)\), we check Appendix Table A.3, from there

\(\phi (1.18) = 0.8810\)

Hence

\(\begin{array}{l}P(4000 < X) = 1 - 0.8810\\{\rm{P}}(4000 < {\rm{X}}) = 0.1190\end{array}\)

Proposition: Let Z be a continuous rv with cdf \(\phi (z)\). Then for any a,

\(P(a < Z) = 1 - \phi (a)\)

The probability that the birth weight of a randomly selected baby of this type is between 3000 and \(4000\;{\rm{g}}\)can be represented as \(P(3000 < X < 4000)\). Standardizing gives:

\(3000 < X < 4000\)

if and only if

\(\begin{array}{c}\frac{{3000 - 3432}}{{482}} < \frac{{X - 3432}}{{482}} < \frac{{4000 - 3432}}{{482}}\\\frac{{ - 432}}{{482}} < \frac{{X - 3432}}{{482}} < \frac{{568}}{{482}}\\ - 0.90 < Z < 1.18\end{array}\)

Thus

\(P(3000 < X < 4000) = P( - 0.90 < Z < 1.18)\)

Here Z is a standard normal distribution rv with cdf \(\phi (z)\). Hence, we can write

\(P(3000 < X < 4000) = \phi (1.18) - \phi ( - 0.9)\)

To get \(\phi (1.18)\)and \(\phi ( - 0.9)\), we check Appendix Table A.3, from there

\(\phi ( - 0.9) = 0.1841{\rm{ and }}\phi (1.18) = 0.8810\)

Hence

\(\begin{array}{l}P(3000 < X < 4000) = 0.8810 - 0.1841\\{\rm{P}}(3000 < {\rm{X}} < 4000) = 0.6969\end{array}\)

Proposition: Let Z be a continuous rv with cdf \(\phi (z)\). Then for any a and b with a<b,

\(P(a < Z < b) = \phi (b) - \phi (a)\)

(b) The probability that the birth weight of a randomly selected baby of this type either less than \(2000\;{\rm{g}}\)or greater than \(5000{\rm{g}}\)can be represented as \(P(X < 2000\,or\,X > 5000)\):

\(P(2000 < X{\rm{ or }}X > 5000) = P(X < 2000) + P(X > 5000)\)

First, we calculate \(P(X > 5000)\). Standardizing gives:

\(5000 < X\)

if and only if

\(\begin{array}{c}\frac{{5000 - 3432}}{{482}} < \frac{{X - 3432}}{{482}}\\\frac{{1568}}{{482}} < \frac{{X - 3432}}{{482}}\\3.25 < Z\end{array}\)

Thus

\(P(5000 < X) = P(3.25 < Z)\)

Here, Z is a standard normal distribution rv with cdf \(\phi (z)\). Hence, we can write

\(P(5000 < X) = 1 - \phi (3.25)\)

To get\(\phi (3.25)\), we check Appendix Table A.3, from there

\(\phi (3.25) = 0.9994\)

Hence,

\(\begin{array}{l}P(5000 < X) = 1 - 0.9994\\P(5000 < X) = 0.0006\end{array}\)

Now we calculate \(P(X < 2000)\). Standardizing gives:

\(X < 2000\)

if and only if

\(\begin{array}{c}\frac{{X - 3432}}{{482}} < \frac{{2000 - 3432}}{{482}}\\\frac{{X - 3432}}{{482}} < \frac{{ - 1432}}{{482}}\\Z < - 2.97\end{array}\)

Thus

\(\begin{array}{l}P(X < 2000) = P(Z < - 2.97)\\P(X < 2000) = \phi ( - 2.97)\\P(X < 2000) = 0.0014\end{array}\)

Using equations (1) and (2), we can write:

\(\begin{array}{l}P(2000 < X{\rm{ or }}X > 5000) = P(X < 2000) + P(X > 5000) = 0.0014 + 0.0006\\P(2000 < X{\rm{ or }}X > 5000) = 0.002\end{array}\)

Proposition: Let Z be a continuous rv with cdf \(\phi (z)\). Then for any a,

\(\begin{array}{l}P(Z < a) = \phi (a)\\P(a < Z) = 1 - \phi (a)\end{array}\)

(c) First, we need to convert the units of given weight. \(7lb\)is equal to \(3175.15g\). The probability that the birth weight of a randomly selected baby of this type exceeds \(7lb\) can be represented as\(P(X > 3175.15)\). Standardizing gives:

\(3175.15 < X\)

if and only if

\(\begin{array}{c}\frac{{3175.15 - 3432}}{{482}} < \frac{{X - 3432}}{{482}}\\\frac{{ - 256.85}}{{482}} < \frac{{X - 3432}}{{482}}\\ - 0.53 < Z\end{array}\)

Thus

\(\begin{array}{c}P(3175.15 < X) = P( - 0.53 < Z)\\ = 1 - \phi ( - 0.53)\;\;\;({\rm{ check Appendix Table A}}{\rm{.3)}}\\ = 1 - 0.2981\\P(3175.15 < X) = 0.7019\end{array}\)

(d) Most extreme \(0.1\% \)values are the highest \(0.05\% \)and lowest \(0.05\% \). The percentile values (z-scores) that separates them are \({Z_{0.0005}}\)and \({Z_{0.9995}}\)respectively. Which means the most extreme values are values whose z-score are greater than \({Z_{0.0005}}\)or lower than \({Z_{0.9995}}\)

If we recall, according to the definition, \({Z_\alpha }\)is the \(100{(1 - \alpha )^{{\rm{th }}}}\)percentile of the standard normal distribution.

\({Z_{0.0005}}\)means that area to the left of \({Z_{0.0005}}\)under standard normal distribution curve is \(0.9995\)

we can also say that :

\(\phi \left( {{z_{0.0005}}} \right) = 0.9995\)

Where, \(\phi (z)\)is the cdf of standard normal distributed rv Z.

We check Appendix Table A.3 to see if \(\phi (z)\)is equal to \(0.9995\)for any z. From there

\({z_{0.0005}} \approx 3.3\)

Using symmetry of standard normal distribution, we can say that:

\({z_{0.9995}} = - {z_{0.0001}} = - 3.3\)

Let \({X_{0.0005}}\)and \({X_{0.9995}}\)denote the values of rv corresponding to z-value of \({Z_{0.0005}}\)and \({Z_{0.9995}}\)respectively. Then

\(\begin{array}{c}\frac{{{X_{0.0005}} - 3432}}{{482}} = {Z_{0.0005}}\\\frac{{{X_{0.0005}} - 3432}}{{482}} = 3.3\\{X_{0.0005}} = 3432 + (482)(3.3)\\{X_{0.0005}} = 5022.6\end{array}\)

Similarly

\(\begin{array}{l}{X_{0.9995}} = 3432 + (482)( - 3.3)\\{X_{0.9995}} = 1841.4\end{array}\)

Hence most extreme \(0.1\% \)weights are either less than \(184.4\;{\rm{g}}\)or more than \(5022.6\;{\rm{g}}\)

(e) The shape of the distribution is not affected if the random variable is multiplied or divided by a constant, hence the shape of distribution of weight expressed in pounds will be bell shaped and normally distributed. First, we recall the following proposition:

Proposition: When \(h(X) = aX + b\), then the expected value and standard deviation of \(h(X)\)satisfy the following properties:

\(\begin{array}{l}E(h(X)) = a\\E(X) + b{\sigma _{h(x)}} = a{\sigma _x}\end{array}\)

Let the birth weights of the babies expressed in pounds be represented by a rv $Y$.

\(Y = \frac{X}{{453.592}}\)

Then mean and standard deviation of $Y$ are given as:

\(\begin{array}{l}{\mu _Y} = \frac{{3432}}{{453.592}} \approx 7.566{\rm{ pounds }}\\{\sigma _Y} = 482453.592 \approx 1.063{\rm{ pounds }}\end{array}\)

The probability that the birth weight of a randomly selected baby of this type exceeds \(7lb\)can be represented as \(P(Y > 7)\). Standardizing gives:

\(7 < Y\)

if and only if

\(\begin{array}{c}\frac{{7 - 7.566}}{{1.063}} < \frac{{X - 7.566}}{{1.063}}\\\frac{{ - 0.566}}{{1.063}} < \frac{{X - 7.566}}{{1.063}}\\ - 0.53 < Z\end{array}\)

Thus

\(\begin{array}{c}P(7 < Y) = P( - 0.53 < Z)\\ = 1 - \phi ( - 0.53)\;\;\;({\rm{ check Appendix Table A}}{\rm{.3 }})\\ = 1 - 0.2981\\P(7 < Y) = 0.7019\end{array}\)

Hence the probability calculated is same as one calculated in part(c)