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Novay_Z [31]
3 years ago
10

One rule of thumb for estimating crowds is that each person occupies 2.3 square feet. Use this value to estimate the size of a c

rowd that is 8 feet wide on both sides of the street standing along a 4.1-mile section of a parade route.
Mathematics
1 answer:
dybincka [34]3 years ago
7 0

Answer:

150,595

Step-by-step explanation:

First, convert miles to feet:

4.1 mi × (5280 ft/mi) = 21648 ft

The area that the crowd takes up is:

2 × 8 ft × 21648 ft = 346368 ft²

The number of people is:

346368 ft² × (1 person / 2.3 ft²) = 150594.8

Rounding to the nearest integer, there are approximately 150,595 people in the crowd.

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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
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Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c)  Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046  

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:                                                

                    n*p = 745*0.037 = 27.565 > 5

                    n*q = 745*0.963 = 717.435 > 5

                    Normal Approximation is valid.

a) P ( X >= 15 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

 - Then the parameters u mean and σ standard deviation for normal distribution are:

                u = n*p = 27.565

                σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

                X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= -2.5358 ) = 0.9944

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

                Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= 0.37556 ) = 0.3182

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

                Bi [P ( X >=30 ) ] ≈ 0.3182  

c) P ( 25=< X =< 35 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

                N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

                P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

               N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

                Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

                N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

               P ( Z > 2.60762 ) = 0.0046

               N [ P ( X > 41 ) ] =  P ( Z > 2.60762 ) = 0.0046

Hence,

                Bi [P ( X >40 ) ] ≈ 0.0046  

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