Chapter 7.2, Problem 50E

(a)

To determine

The 40^th percentile of the diameter.

Answer to Problem 50E

The 40^th percentile of the diameter is $72.97$

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}\mu =75\\ \sigma =8\end{array}$

Probability density function of normal distribution is $f(x,\sigma ,\mu )=\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}:-\infty <x<\infty$

Now, for determining 60^th percentile of the diameter,

  ${\displaystyle \underset{-\infty }{\overset{x}{\int }}\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}=}0.4$

Using calculator, $x=72.97$

(b)

To determine

The 65^th percentile of the diameter.

Answer to Problem 50E

The 65^th percentile of the diameter is $78.08$

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}\mu =75\\ \sigma =8\end{array}$

Probability density function of normal distribution is $f(x,\sigma ,\mu )=\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}:-\infty <x<\infty$

Now, for determining 32^th percentile of the diameter,

  ${\displaystyle \underset{-\infty }{\overset{x}{\int }}\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}=}0.65$

Using calculator, $x=78.08$

(c)

To determine

The minimum score above which top $10\%$ of students will lie

Answer to Problem 50E

The minimum score above which top $10\%$ of students will lie is $85.25$

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}\mu =75\\ \sigma =8\end{array}$

Probability density function of normal distribution is $f(x,\sigma ,\mu )=\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}:-\infty <x<\infty$

For determining the minimum score above which top $10\%$ of students will lie, one has to find 90^th percentile of the score of statistics exam.

Now,

  ${\displaystyle \underset{-\infty }{\overset{x}{\int }}\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}=}0.90$

Using calculator, $x=85.25$

The minimum score above which top $10\%$ of students will lie is $85.25$

(d)

To determine

The range in which middle $60\%$ of the diameters will lie.

Answer to Problem 50E

Between $68.26$ $25.04$ and $81.73$ middle $60\%$ of the scores of statistics exam will lie.

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}\mu =25.1\\ \sigma =0.08\end{array}$

Probability density function of normal distribution is $f(x,\sigma ,\mu )=\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}:-\infty <x<\infty$

For determining the range in which middle $60\%$ of the scores will lie, one has to find 20^thpercentile from left side and 20^th percentile from right side.

For left 25^th percentile,

  ${\displaystyle \underset{-\infty }{\overset{x}{\int }}\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}=}0.2$

Using calculator, $x=68.26$

For right 25^th percentile,

  ${\displaystyle \underset{x}{\overset{\infty }{\int }}\frac{1}{\sqrt{2\pi }\sigma }{e}^{\frac{-1}{2}{\left(x-\mu \right)}^2}=}0.2$

Using calculator, $x=81.73$

Now, between $68.26$ and $81.73$ middle $60\%$ of the scores of statistics exam will lie