Chapter 6.4, Problem 28E

To determine

a.

To find:

The probability that a randomly selected student will drink less than suggested 40.5 oz of water in a day supposing the amounts of water that schoolchildren actually consume in a day are approximately normally distributed with a mean of 32.0 oz and a standard deviation of 7.1 oz.

Answer to Problem 28E

Solution:

The probability that a randomly selected student will drink less than suggested water is 0.884.

Explanation of Solution

Given:

Mean $\mu =32.0$

Standard deviation $\sigma =7.1$

Suggested intake according to report $X_s=40.5$

Cut-off intake $X_c=36$

Percentile $a=5\%$

Description:

In statistics, the normal Gaussian distribution with the following characteristics:

$\mu =0$, and

$\sigma =1$

is called standard normal distribution with probability density function:

$f\left(x|0,1\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$

When this function is plotted on the $x$-axis, the area between the curve and $x$-axis is equal to one, because the sum of probabilities is unity.

Area on the left of a given $z$-value (say $z=a$) is the probability $P\left(X\le a\right)$ given by the integration:

${\displaystyle \underset{-\infty }{\overset{a}{\int }}f\left(x\right)dx}$,

which can be obtained from the normal distribution table.

The $x$-value, $z$-value, the mean, along with the standard deviation are related as below:

$z=\frac{x-\mu }{\sigma }$,

or equivalently

$x=z\times \sigma +\mu$.

Now,

In the first subpart, in order to find the probability the table is consulted to look-up the value in the row denoted by unit and tenth digits and the column by hundredth digit before which the $z$-value $z_s$ is calculated.

Calculation:

In the first part, using the given data and above relation:

$z_s=\frac{X_s-\mu }{\sigma }$

that is:

$z_s=\frac{40.5-32.0}{7.1}$

which gives:

$z_s=1.197$.

This $z$-value lies between 1.19 and 1.20.

$\text{Area on the left}=\frac{0.88298+0.88493}{2}$

which is:

$\text{Area on the left}=0.883955$.

Rounding off to three places after the decimal, the probability is $0.884$ as $9\ge 5$.

Conclusion:

The probability that a randomly selected student will drink less than suggested water is 0.884.

To determine

b.

To find:

Amount by which the mean needs to be increased so that only $5\%$ of students drink less than 36 oz per week, if the standard deviation remained the same and the daily water intakes were still normally distributed.

Answer to Problem 28E

Solution:

Amount by which the mean needs to be increased so that only $5\%$ of students drink less than 36 oz per week is $15.68$ oz.

Explanation of Solution

Given:

Mean $\mu =32.0$

Standard deviation $\sigma =7.1$

Suggested intake according to report $X_s=40.5$

Cut-off intake $X_c=36$

Percentile $a=5\%$

Description:

In statistics, the normal Gaussian distribution with the following characteristics:

$\mu =0$, and

$\sigma =1$

is called standard normal distribution with probability density function:

$f\left(x|0,1\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$

When this function is plotted on the $x$-axis, the area between the curve and $x$-axis is equal to one, because the sum of probabilities is unity.

Area on the left of a given $z$-value (say $z=a$) is the probability $P\left(X\le a\right)$ given by the integration:

${\displaystyle \underset{-\infty }{\overset{a}{\int }}f\left(x\right)dx}$,

which can be obtained from the normal distribution table.

The $x$-value, $z$-value, the mean, along with the standard deviation are related as below:

$z=\frac{x-\mu }{\sigma }$,

or equivalently

$x=z\times \sigma +\mu$.

Now,

In the second part, the appropriate $z$-value $z*$ is first obtained from the percentile given by converting percentile to area on left after which it is used to calculate the new mean by:

$\mu *=X_c-z*\times \sigma$.

Thereafter, the required increase in mean is computed.

Calculation:

In the second part, area on the left is equal to the decimal representation of percentile, that is:

$\text{Area}\text{on}\text{left}=0.05$.

From the table, the $z$-value is obtained at $z*=-1.645$ which is between $-1.64$ and $-1.65$.

Using the relation defined above,

$\mu *=X_c-z*\times \sigma$,

the new mean is found as:

$\mu *=36-\left(-1.645\right)\times 7.1$

which gives the new mean:

$\mu *=47.6795$.

Previous mean $\mu =32$, so the increase in mean required is:

$47.6795-32=15.6795$.

Since thousandth place digit is 9 and $9\ge 5$, thus rounding off the difference till the hundredth place digit is:

$\text{difference}=15.68$.

Conclusion:

Amount by which the mean needs to be increased so that only $5\%$ of students drink less than 36 oz per week is $15.68$ oz.