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Zarrin [17]
3 years ago
11

. Suppose the weight of Chipotle burritos follows a normal distribution with mean of 450 grams, and variance of 100 grams2 . Def

ine 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
Mathematics
1 answer:
mylen [45]3 years ago
6 0

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

Now from the normal distribution table  the value for P (Z  is  

      P(X  <  445  ) =  P (Z

=>   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

       

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