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attashe74 [19]
3 years ago
9

52% of all Americans are home owners. If 45 Americans are randomly selected, find the probability that a. Exactly 21 of them are

are home owners. b. At most 24 of them are are home owners. c. At least 23 of them are home owners. d. Between 20 and 24 (including 20 and 24) of them are home owners.
Mathematics
1 answer:
Nataly [62]3 years ago
8 0

Answer:

0.0918 ; 0.6277 ; 0.6064; 0.5055

Step-by-step explanation:

Given that :

Proportion of home owners : p = 0.52

Number of trials = 45

Using binomial distribution formula :

P(x =x) = nCr * p^x * (1 - p)^(n-x)

A.) probability that exactly 21 are homeowners :

P(x = 21) = 45C21 * 0.52^21 * (1 - 0.52)^(45 - 21)

P(x = 21) = 45C21 * 0.52^21 * 0.48^24

P(x = 21) = 3773655750150 * 0.52^21 * 0.48^24

P(x = 21) = 0.0918

b. At most 24 of them are are home

Using the binomial probability to save computation time.

P(x ≤ 24) = p(0) + p(1) +.... + p(24)

P(x ≤ 24) = 0.6277

C.) At least 23 of them are home owners.

P(x ≥ 23) = p(23) + p(24) +... + p(45)

P(x ≥ 23) = 0.6064

d. Between 20 and 24 (including 20 and 24) of them are home owners.

P(20 ≤ x ≤ 24) = p(20) + p(21) + p(22) + p(23) + p(24)

P(20 ≤ x ≤ 24) = 0.0711 + 0.0918 + 0.1084 + 0.1175 + 0.1167

P(20 ≤ x ≤ 24) = 0.5055

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Answer:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

\bar X=19.87 represent the sample mean

\sigma=0.4 represent the population deviation

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\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

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Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:\mu \geq 20  

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The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing the info given we got:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

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Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

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