Question

A survey found that​ women's heights are normally distributed with mean 64.2 in. and standard deviation 4.2 in. The survey also found that​ men's heights are normally distributed with mean 70.1 in. and standard deviation 4.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in. Complete parts​ (a) and​ (b) below

Answer 1

A survey found that​ women's heights are normally distributed with mean 64.2 in. and standard deviation 4.2 in. The survey also found that​ men's heights are normally distributed with mean 70.1 in. and standard deviation 4.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in. Complete parts​ (a) and​ (b) below

a) find the percentage of women meeting the height requirement

 

b) find the percentage of men meeting the height requirement

 

Answer:

a

P(57 <  X <  63) = 34.4%

b

P(57 <  X <  63) =  4.096%

Step-by-step explanation:

From the question we are told that

    The mean of women's height is $\mu_ 1 = 64.2$

    The standard deviation is $\sigma_1 = 4.2 \ in$

     The mean of men' height is $\mu_2 = 70.1 \ in$

     The standard deviation of men's height is $\sigma_2 = 4.1 \ in$

      The minimum height requirement is  a =  57 in

      The maximum height requirement is  b = 63 in

Generally the percentage of women meeting the height requirement  is mathematically represented as

     $P(57 < X < 63) = P(\frac{ 57 - 64.2}{ 4.2} < \frac{X - \mu_1}{\sigma_1} < \frac{ 63 - 64.2}{ 4.2} )$

$\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )$

So

      $P(57 < X < 63) = P( -1.7143 < Z <-0.2857)$

=>  $P(57 < X < 63) = P ( Z <-0.2857) - P(Z < -1.7143)$

Generally from the z table the probability of  ( Z <-0.2857)  and  (Z <  -1.7143) is  

    P( Z <-0.2857)   =  0.38755

 and

    P(Z <  -1.7143) =  0.043237

So

   P(57 <  X <  63) = 0.38755  - 0.043237

=>  P(57 <  X <  63) = 0.3443

Converting to percentage

     P(57 <  X <  63) = 0.3443  * 100 =  34.4%

Generally the percentage of men meeting the height requirement  is mathematically represented as

     $P(57 < X < 63) = P(\frac{ 57 - 70.1}{ 4.1} < \frac{X - \mu_1}{\sigma_1} < \frac{ 63 - 70.1}{ 4.1} )$

$\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )$

So

      P(57 <  X <  63) = P( -3.1951  < Z <-1.7317)

=>  $P(57 < X < 63) = P ( Z <-1.7317) - P(Z < -3.1951)$

Generally from the z table the probability of  ( Z <-0.2857)  and  (Z <  -1.7143) is  

    P( Z <-1.7317)   =  0.041663

 and

    P(Z <  -3.1951 ) =  0.00069891

So

   P(57 <  X <  63) = 0.041663  - 0.00069891

=>  P(57 <  X <  63) = 0.04096

Converting to percentage

     P(57 <  X <  63) = 0.04096   * 100 =  4.096%