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Greeley [361]
3 years ago
9

A distribution of measurements is relatively mound-shaped with a mean of 40 and a standard deviation of 15. Use this information

to find the proportion of measurements in the given interval. between 25 and 55
Mathematics
1 answer:
MArishka [77]3 years ago
4 0

Answer:

The proportion of measurements between 25 and 55

P( 25 ≤ X≤ 55) = 0.6826

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 40

Given that standard deviation of the Population = 15

Let 'x' be the random variable in normal distribution

Let 'X' = 25

Z = \frac{x-mean}{S.D} = \frac{25-40}{15} = -1

Let 'X' = 55

Z = \frac{x-mean}{S.D} = \frac{55-40}{15} = 1

<u><em>Step(ii):-</em></u>

The probability that between 25 and 55

P( 25 ≤ X≤ 55) = P( -1≤z≤1)

                       = A(1) - A(-1)

                      = A(1) + A(1)

                     = 2 × A(1)

                    = 2× 0.3413

                   = 0.6826

The proportion of measurements between 25 and 55

P( 25 ≤ X≤ 55) = 0.6826

<u><em>Final answer:-</em></u>

The proportion of measurements between 25 and 55

P( 25 ≤ X≤ 55) = 0.6826

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The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 1252
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Answer:

a)  P (  1100 < X < 1400 ) = 0.755

b) P (  X < 1000 ) = 0.755

c) proportion ( X > 1200 ) = 65.66%

d) 5.87% percentile

Step-by-step explanation:

Solution:-

- Denote a random variable X: The number of chocolate chip in an 18-ounce bag of chocolate chip cookies.

- The RV is normally distributed with the parameters mean ( u ) and standard deviation ( s ) given:

                               u = 1252

                               s = 129

- The RV ( X ) follows normal distribution:

                       X ~ Norm ( 1252 , 129^2 )  

a) what is the probability that a randomly selected bag contains between 1100 and 1400 chocolate​ chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

     P ( x1 < X < x2 ) = P ( [ x1 - u ] / s < Z <  [ x2 - u ] / s )

- Taking the limits x1 = 1100 and x2 = 1400. The standard normal values are:

     P (  1100 < X < 1400 ) = P ( [ 1100 - 1252 ] / 129 < Z <  [ 1400 - 1252 ] / 129 )

                                        = P ( - 1.1783 < Z < 1.14728 )

       

- Use the standard normal tables to determine the required probability defined by the standard values:

       P ( -1.1783 < Z < 1.14728 ) = 0.755

Hence,

      P (  1100 < X < 1400 ) = 0.755   ... Answer

b) what is the probability that a randomly selected bag contains fewer than 1000 chocolate​ chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

     P ( X < x2 ) = P ( Z <  [ x2 - u ] / s )

- Taking the limit x2 = 1000. The standard normal values are:

     P (  X < 1000 ) = P ( Z <  [ 1000 - 1252 ] / 129 )

                                        = P ( Z < -1.9535 )

       

- Use the standard normal tables to determine the required probability defined by the standard values:

       P ( Z < -1.9535 ) = 0.0254

Hence,

       P (  X < 1000 ) = 0.755   ... Answer

​(c) what proportion of bags contains more than 1200 chocolate​ chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

     P ( X > x1 ) = P ( Z >  [ x1 - u ] / s )

- Taking the limit x1 = 1200. The standard normal values are:

     P (  X > 1200 ) = P ( Z >  [ 1200 - 1252 ] / 129 )

                                        = P ( Z > 0.4031 )

       

- Use the standard normal tables to determine the required probability defined by the standard values:

       P ( Z > 0.4031 ) = 0.6566

Hence,

      proportion of X > 1200 = P (  X > 1200 )*100 = 65.66%   ... Answer

d) what is the percentile rank of a bag that contains 1050 chocolate​ chips?

- The percentile rank is defined by the proportion of chocolate less than the desired value.

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

     P ( X < x2 ) = P ( Z <  [ x2 - u ] / s )

- Taking the limit x2 = 1050. The standard normal values are:

     P (  X < 1050 ) = P ( Z <  [ 1050 - 1252 ] / 129 )

                                        = P ( Z < 1.5659 )

       

- Use the standard normal tables to determine the required probability defined by the standard values:

       P ( Z < 1.5659 ) = 0.0587

Hence,

       Rank = proportion of X < 1050 = P (  X < 1050 )*100

                 = 0.0587*100 %  

                 = 5.87 % ... Answer

6 0
3 years ago
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