Question

The mean of a normal distribution is 400 pounds. The standard deviation is 10 pounds. What is the probability of a weight between 415 pounds and the mean of 400 pounds

Answer 1

Answer:

The probability is $P(x_1 \le X \le x_2 ) = 0.4332$

Step-by-step explanation:

From the question we are told that

   The population mean is  $\mu = 400$

   The  standard deviation is  $\sigma = 10$

   The considered values are  $x_1 = 400 \to x_2 = 415$

Given that the weight follows a normal distribution

     i.e       $\approx X (\mu , \sigma )$

Now the probability of a weight between 415 pounds and the mean of 400 pounds is mathematically as

     $P(x_1 \le X \le x_2 ) = P(\frac{x_1 - \mu }{\sigma } \le \frac{X - \mu }{\sigma } \le \frac{x_2 - \mu }{\sigma } )$

So  $\frac{X - \mu }{\sigma }$ is equal to Z (the standardized value of  X  )

Hence we have  

     $P(x_1 \le X \le x_2 ) = P(\frac{x_1 - \mu }{\sigma } \le Z \le \frac{x_2 - \mu }{\sigma } )$

substituting values

      $P(x_1 \le X \le x_2 ) = P(\frac{400 - 400 }{10 } \le Z \le \frac{415 - 400}{415 } )$

      $P(x_1 \le X \le x_2 ) = P(0\le Z \le 1.5 )$

      $P(x_1 \le X \le x_2 ) = P( Z < 1.5) - P( Z < 0)$

From the standardized normal distribution table  $P( Z< 1.5) = 0.9332$ and

   $P( Z < 0) = 0.5$

So

     $P(x_1 \le X \le x_2 ) = 0.9332 - 0.5$

     $P(x_1 \le X \le x_2 ) = 0.4332$

NOTE :  This above  values obtained from the standardized normal distribution table can also be obtained using the P(Z) calculator at  (calculator dot net).