Question

. Suppose the weight of Chipotle burritos follows a normal distribution with mean of 450 grams, and variance of 100 grams2 . Define a random variable to be the weight of a randomly chosen burrito. (a) What is the probability that a Chipotle burrito weighs less than 445 grams? (3 points) (b) 20% of Chipotle burritos weigh more than what weig

Answer 1

Complete Question

Suppose the weight of Chipotle burritos follows a normal distribution with mean of 450 grams, and variance of 100 grams2 . Define a random variable to be the weight of a randomly chosen burrito.

(a) What is the probability that a Chipotle burrito weighs less than 445 grams? (3 points)

(b) 20% of Chipotle burritos weigh more than what weight

Answer:

a

   $P(X < 445 )= 0.3085$

b

  $k = 458.42$

Step-by-step explanation:

From question we are told that

     The population mean is $\mu = 450 \ g$

      The variance is $var = 100 \ g^2$

      The  consider weight is  $x = 445 \ g$

The  standard deviation is mathematically represented as

     $\sigma = \sqrt{var}$

substituting values

     $\sigma = \sqrt{ 100}$

     $\sigma = 10$

Given that weight of Chipotle burritos follows a normal distribution the  the probability that a Chipotle burrito weighs less than x grams is mathematically represented as

        $P(X < x ) = P ( \frac{X - \mu }{\sigma } < \frac{x - \mu }{\sigma } )$

Where  $\frac{X - \mu }{\sigma }$ is  equal to z (the standardized values of the random number X )

So

     $P(X < x ) = P (Z < \frac{x - \mu }{\sigma } )$

substituting values

     $P(X < 445 ) = P (Z < \frac{445 - 450 }{10} )$

      $P(X < 445 ) = P (Z <-0.5 )$

Now from the normal distribution table  the value for $P (Z <-0.5 )$  is  

      $P(X < 445 ) = P (Z <-0.5 ) = 0.3085$

=>   $P(X < 445 )= 0.3085$

Let the  probability  of the Chipotle burritos weighting more that k be 20% so  

       $P(X > k ) = P ( \frac{X - \mu }{\sigma } > \frac{k - \mu }{\sigma } ) = 0.2$

=>    $P (Z> \frac{k - \mu }{\sigma } ) = 0.2$

=>    $P (Z> \frac{k - 450}{10 } ) = 0.2$

From the normal distribution table the value of z  for  $P (Z> \frac{k - \mu }{\sigma } ) = 0.2$ is  

    $z = 0.8416$

=>   $\frac{k - 450}{10 } = 0.8416$

=>   $k = 458.42$