Loose Leaf Version For Elementary Statistics
Loose Leaf Version For Elementary Statistics
3rd Edition
ISBN: 9781260373523
Author: William Navidi Prof., Barry Monk Professor
Publisher: McGraw-Hill Education
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Chapter 6.2, Problem 39E

High blood pressure: The National Health and Nutrition Survey reported that 30% of adults in the United States have hypertension (high blood pressure). A sample of 25 adults is studied.

What is the probability that exactly 6 of them have hypertension?

What is the probability that more than 8 have hypertension?

What is the probability that fewer than 4 have hypertension?

Would it be unusual if more than 10 of them have hypertension?

What is the mean number who have hypertension in a sample of 25 adults?

What is the standard deviation of the number who have hypertension in a sample of 25 adults?

(a)

Expert Solution
Check Mark
To determine

To find: The probability that exactly 6 have hypertension.

Answer to Problem 39E

The probability that exactly 6 have hypertensionis P(6)=0.147166 .

Explanation of Solution

Given:

  n=25

  p=0.30

Adults having hypertension in US- 30%

Calculation:

Here, n=25 p=0.30 x=6

  q=1p=10.30=0.7

Probability of binomial distribution is,

  P(x)=Cnxpxqnx P(6)= C 25 6 (0.30) 6 (0.7) 256 P(6)=0.147166

Hence, theprobability that exactly 6 have hypertensionis P(6)=0.147166 .

Conclusion:

Therefore, the probability that exactly 6 have hypertensionis P(6)=0.147166 .

(b)

Expert Solution
Check Mark
To determine

To find: The probability that more than 8 have hypertension.

Answer to Problem 39E

The probability that more than 8 have hypertensionis 0.323054

Explanation of Solution

Calculation:

Here, n=25 p=0.30 x<8

  q=1p=10.30=0.7

  P(more than 8)=P(9)+P(10)+P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)+                                            P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)                                           

Probability of binomial distribution is,

   P(x)= C n x p x q nx

   P(9)= C 25 9 (0.30) 9 (0.7) 259

   P(9)=0.133636

   P(10)= C 25 10 (0.30) 10 (0.7) 2510

   P(10)=0.0916360

   P(11)= C 25 11 (0.30) 11 (0.7) 2511

   P(11)=0.0535535

   P(12)= C 25 12 (0.30) 12 (0.7) 2512

   P(12)=0.0267768

   P(13)= C 25 13 (0.30) 13 (0.7) 2513

   P(13)=0.0114758

   P(14)= C 25 14 (0.30) 14 (0.7) 2514

   P(14)=0.0042156

   P(15)= C 25 15 (0.30) 15 (0.7) 2515

   P(15)=0.0013249

   P(16)= C 25 16 (0.30) 16 (0.7) 2516

   P(16)=0.0003549

   P(17)= C 25 17 (0.30) 17 (0.7) 2517

   P(17)=0.0000805

   P(18)= C 25 18 (0.30) 18 (0.7) 2518

   P(18)=0.0000153

   P(19)= C 25 19 (0.30) 19 (0.7) 2519

   P(19)=0.0000024

   P(20)= C 25 20 (0.30) 20 (0.7) 2520

   P(20)=0.00000

   P(21)= C 25 21 (0.30) 21 (0.7) 2521

   P(21)=0.00000

   P(22)= C 25 22 (0.30) 22 (0.7) 2522

   P(22)=0.00000

   P(23)= C 25 23 (0.30) 23 (0.7) 2523

   P(23)=0.00000

   P(24)= C 25 24 (0.30) 24 (0.7) 2524

   P(24)=0.00000

   P(25)= C 25 25 (0.30) 25 (0.7) 2525

   P(25)=0.00000

  P(more than 8)=P(9)+P(10)+P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)+                                            P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)=0.133636+0.0916360+0.0535535+0.0267768+0.0114758+0.0042156+0.0013249+0.0003549                0.0000805+0.0000153+0.0000024+0.00000+0.00000+0.00000+0.00000+0.00000+0.00000=0.323054                                       

Hence, theprobability that more than 8 have hypertensionis 0.323054

Conclusion:

Therefore, the probability that more than 8 have hypertensionis 0.323054

(c)

Expert Solution
Check Mark
To determine

To find: The probability that fewer than 4 have hypertension.

Answer to Problem 39E

The probability that fewer than 4 have hypertensionis 0.03324       

Explanation of Solution

Calculation:

Here, n=25 p=0.30 x>4

  q=1p=10.30=0.7

  P(fewer than 4)=P(0)+P(1)+P(2)+P(3)                                           

Probability of binomial distribution is,

  Loose Leaf Version For Elementary Statistics, Chapter 6.2, Problem 39E

  P(fewer than 4)=P(0)+P(1)+P(2)+P(3)                           =0.0001341+0.0014369+0.0073896+0.0242800                          =0.03324                                                         

Hence, theprobability that fewer than 4 have hypertensionis 0.03324       

Conclusion:

Therefore, the probability that fewer than 4 have hypertensionis 0.03324       

(d)

Expert Solution
Check Mark
To determine

To find: Whether it is unusual if more than 10 have hypertension.

Answer to Problem 39E

It is not unusual if more than 10 have hypertension.

Explanation of Solution

Calculation:

Here, n=25 p=0.30 x<10

  q=1p=10.30=0.7

  P(more than 10)=P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)+                                    P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)                                        

   P(11)= C 25 11 (0.30) 11 (0.7) 2511

   P(11)=0.0535535

   P(12)= C 25 12 (0.30) 12 (0.7) 2512

   P(12)=0.0267768

   P(13)= C 25 13 (0.30) 13 (0.7) 2513

   P(13)=0.0114758

   P(14)= C 25 14 (0.30) 14 (0.7) 2514

   P(14)=0.0042156

   P(15)= C 25 15 (0.30) 15 (0.7) 2515

   P(15)=0.0013249

   P(16)= C 25 16 (0.30) 16 (0.7) 2516

   P(16)=0.0003549

   P(17)= C 25 17 (0.30) 17 (0.7) 2517

   P(17)=0.0000805

   P(18)= C 25 18 (0.30) 18 (0.7) 2518

   P(18)=0.0000153

   P(19)= C 25 19 (0.30) 19 (0.7) 2519

   P(19)=0.0000024

   P(20)= C 25 20 (0.30) 20 (0.7) 2520

   P(20)=0.00000

   P(21)= C 25 21 (0.30) 21 (0.7) 2521

   P(21)=0.00000

   P(22)= C 25 22 (0.30) 22 (0.7) 2522

   P(22)=0.00000

   P(23)= C 25 23 (0.30) 23 (0.7) 2523

   P(23)=0.00000

   P(24)= C 25 24 (0.30) 24 (0.7) 2524

   P(24)=0.00000

   P(25)= C 25 25 (0.30) 25 (0.7) 2525

   P(25)=0.00000

  P(more than 10)=P(11)+P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)+                                            P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)=0.0535535+0.0267768+0.0114758+0.0042156+0.0013249+0.0003549                0.0000805+0.0000153+0.0000024+0.00000+0.00000+0.00000+0.00000+0.00000+0.00000=0.097782                                       

The obtained probability is not so low. Hence, it is not unusual if more than 10 have hypertension.

Conclusion:

Therefore, itis not unusual if more than 10 have hypertension.

(e)

Expert Solution
Check Mark
To determine

To find: The mean value.

Answer to Problem 39E

The mean valueis 7.5 .

Explanation of Solution

Calculation:

To calculate the mean value for persons having hypertension is computed below.

  μx=np   =25(0.30)   =7.5

Hence, the mean valueis 7.5 .

Conclusion:

Therefore, the mean valueis 7.5 .

(f)

Expert Solution
Check Mark
To determine

To find: The standard deviation.

Answer to Problem 39E

The standard deviation valueis 2.29 .

Explanation of Solution

Calculation:

To calculate the standard deviation value for persons having hypertension is computed below.

  σx=np( 1p)   =25( 0.30)( 10.30)   =2.29

Hence, the standard deviation valueis 2.29 .

Conclusion:

Therefore, the standard deviation valueis 2.29 .

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Chapter 6 Solutions

Loose Leaf Version For Elementary Statistics

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