Question

The mean weight of $500$ male students in a certain college is $151$ pounds and the standard deviation is $15$ pounds. Assuming the weights are normally distributed, find the approximate number of students weighing
(i) between$120$ and $155$ pounds,

Z	0.2667	2.067	2.2667
Area	0.1026	0.4803	0.4881

(ii) more than $185$ pounds.

Answer 1

Hint: To find the approximate number of students weighing between $120$ and$155$ pounds, and more than $185$ pounds, we will first find normal distribution, $Z = \dfrac{{X - \mu }}{\sigma }$ for $X = 120,155,185$ . To find number of students weighing between $120$ and $155$ pounds, first we will find the probability of this using $P(120 < X < 155) = P( - 2.067 < Z < 0.267) = \int\limits_{ - 2.067}^{0.267} {\phi (z)dz} $ . Then multiply the answer with $500$ to get the number of students weighing between $120$ and $155$ . Similarly, we will do for more than $185$ pounds.

Complete step by step answer:
We need to find the approximate number of students weighing between $120$ and$155$ pounds, and more than $185$ pounds.
We know that normal distribution, $Z = \dfrac{{X - \mu }}{\sigma }$ , where $\mu $ is the mean and $\sigma $ is the standard deviation of $X$ .
Given that $\mu = 151$ and $\sigma = 15$ .
Substituting in the normal distribution, we get
$Z = \dfrac{{X - 151}}{{15}}$
(i) When $X = 120$ ,
$Z = \dfrac{{120 - 151}}{{15}} = - 2.067$
When $X = 155$ ,
$Z = \dfrac{{155 - 151}}{{15}} = 0.267$
Therefore, probability of students weighing between$120$ and $155$ pounds is
$P(120 < X < 155) = P( - 2.067 < Z < 0.267) = \int\limits_{ - 2.067}^{0.267} {\phi (z)dz} $
We know that $\int\limits_a^b {f(x)dx} = \int\limits_a^c {f(x)dx} + \int\limits_c^b {f(x)dx} ,a < c < b$
Therefore, $\int\limits_{ - 2.067}^{0.267} {\phi (z)dz} = \int\limits_{ - 2.067}^0 {\phi (z)dz} + \int\limits_0^{0.267} {\phi (z)dz} $
We know that $\int\limits_a^b {f(x)dx} = - \int\limits_b^a {f(x)dx} $
Therefore,
$\int\limits_{ - 2.067}^{0.267} {\phi (z)dz} = - \int\limits_0^{ - 2.067} {\phi (z)dz} + \int\limits_0^{0.267} {\phi (z)dz} $
Now substituting the values from the given table, we get
$\int\limits_{ - 2.067}^{0.267} {\phi (z)dz} = - ( - 0.4803) + 0.1026 = 0.5829$
Now, given that the total number of male students.$N = 500$ .
Hence, total number of students whose weights are in between $120$ and$155$ pounds is
$0.5829 \times 500 = 291$
(ii) We have normal distribution, $Z = \dfrac{{X - 151}}{{15}}$
We need to find the number of students weighing more than $185$ pounds.
So, when $X = 185$ ,
$Z = \dfrac{{185 - 151}}{{15}} = 2.2667$
Probability of students weighing more than $185$ pounds is
$P(X > 120) = \int\limits_{2.2667}^\infty {\phi (z)dz} $
This can be written as
$P(X > 120) = \int\limits_{2.2667}^0 {\phi (z)dz} + \int\limits_0^\infty {\phi (z)dz} $
We know that $\int\limits_a^b {f(x)dx} = - \int\limits_b^a {f(x)dx} $
Therefore,
$P(X > 120) = - \int\limits_0^{2.2667} {\phi (z)dz} + \int\limits_0^\infty {\phi (z)dz} $
By substituting the values, we get
$P(X > 120) = - 0.4881 + 0.5 = 0.0119$
Therefore, the total number of students weighing more than $185$ pounds is $500 \times 0.0119 = 5.95 \approx 6$ .

Note: We will be using normal distribution, $Z = \dfrac{{X - \mu }}{\sigma }$ to solve problems with mean and standard deviation. The properties of definite integrals are also used and hence you must be thorough with it. Be careful in taking the intervals of the integral.