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3 years ago

Bob discovers a box with 30 identical, used, rechargeable batteries in his attic.

Mathematics

1 answer:

monitta3 years ago

5 0

Answer:

a) All of them are out of charge = 9.31x10⁻¹⁰

b) 20% of them are out of charge = 5.529x10⁻⁴

Step-by-step explanation:

This problem can be modeled as a binomial distribution since

There are n repeated trials and all of them are independent of each other.

There are only two possibilities: battery is out of charge and battery is not out of charge.

The probability of success does not change with trial to trial.

Since it is given that it is equally likely for the battery to be out of charge or not out of charge so probability of success is 50% or 0.50

P = 0.50

1 - P = 0.50

a) All of them are out of charge?

Probability = nCx * P^x * (1 - P)^n-x

Probability = ₃₀C₃₀(0.50)³⁰(0.50)⁰

Probability = 9.31x10⁻¹⁰

b) 20% of them are out of charge?

0.20*30 = 6 batteries are out of charge

Probability =₃₀C₆(0.50)²⁴(0.50)⁶

Probability = 5.529x10⁻⁴

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Find answers below

Step-by-step explanation:

H0: P <= 0.5

Ha: P > 0.5

, the number who prefer gut strings is <= a number or the test tends towards the left-tailed.

{0,1,2,3,4,5} ;

{15,16,17,18,19,20} is a right-tailed test and not appropriate for H0:

{ 0,1,2,3,17,18,19,20} is two-tailed and not appropriate for H0:

Does the region specify a level .05 test? No

P = proportion who prefer gut strings to nylon

P = X /20

Assume alpha = 0.05

z(alpha) = -1.645

Reject if (x/20 - 0.5) / sqrt[ (0.5)(0.5)/20 ] < -1.645

Reject if (x/20 - 0.5) < < (-1.645) sqrt ( (0.5)(0.5)/20 )

Reject if x/20 < (-1.645) sqrt ( (0.5)(0.5)/20 ) + 0.5

Reject if x/20 < 0.316

Reject if x < (0.316)(20) = 6.32

{0,1,2,3,4,5,6} is the region for the best level 0.05 test

According to (a), reject H0 if x <= 5

P( Type II error) = P( do not reject H0/ when Ha is true)

P( Type II error) = P( x > 5/ P=0.6)

x ---p(x)

6 0.004854 0.998388

7 0.014563

8 0.035497

9 0.070995

10 0.117142

11 0.159738

12 0.179706

13 0.165882

14 0.124412

15 0.074647

16 0.034991

17 0.012350

18 0.003087

19 0.000487

20 0.000037

add: 0.9984 -- proba bility of a type II error

Assuming P=0.8

P( Type II error) = P( x > 5/ P=0.8)

6 0.000002 1.000000

7 0.000013

8 0.000087

9 0.000462

10 0.002031

11 0.007387

12 0.022161

13 0.054550

14 0.109100

15 0.174560

16 0.218199

17 0.205364

18 0.136909

19 0.057646

20 0.011529

add: 1.0000 probability of a type II error

P( x <= 13) =

0 0.000001

1 0.000019

2 0.000181

3 0.001087

4 0.004621

5 0.014786

6 0.036964

7 0.073929

8 0.120134

9 0.160179

10 0.176197

11 0.160179

12 0.120134

13 0.073929

add: 0.9423 < 0.10 , H0 cannot be rejected

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