Mathematics, 08.03.2021 19:40 oof78

The same survey database cited in exercise 4.3.1 (A-5) shows that 32 percent of U. S. adults indicated that they have been tested for HIV at some point in their life. Consider a simple random sample of 15 adults selected at a time. Let X be the number of adults who have been tested for HIV in the sample. For the following, find the numerical answer and describe the answer in words: a. Three
b. Between five and nine, inclusive
c. More than five, but less than 10
d. Six or more
e. Less than five
f. Find the mean and the variance of the number of people tested for HIV in samples of size 15.

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Answer from: lejeanjamespete1

a. P(X = 3) = 0.145736

b. P(5≤X≤9) = 0.5462

c. P(5<X<10) = 0.3332

d. P(X≥6) = 0.3393

e. P(X<5) = 0.4477

f. Mean = 4.8, Variance = 3.264

Step-by-step explanation:

Given - The same survey database cited in exercise 4.3.1 (A-5) shows

that 32 percent of U.S. adults indicated that they have been

tested for HIV at some point in their life. Consider a simple

random sample of 15 adults selected at a time. Let X be the

number of adults who have been tested for HIV in the sample.

To find - For the following, find the numerical answer and describe

the answer in words:

a. Three

b. Between five and nine, inclusive

c. More than five, but less than 10

d. Six or more

e. Less than five

f. Find the mean and the variance of the number of people

tested for HIV in samples of size 15.

Proof -

Given that , n = 15, p = 32% = 0.32

Now,

P(X = 3) = binomial distribution (3, 15, 0.32, 0)

= $\left(\begin{array}{ccc}15\\3\end{array}\right)(0.32)^{3}(1-0.32)^{15 - 3}$

= 455(0.032768)(0.009774)

= 0.145736

⇒P(X = 3) = 0.145736

P(5≤X≤9) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

= $\left(\begin{array}{ccc}15\\5\end{array}\right)(0.32)^{5}(1-0.32)^{15 - 5}$ + $\left(\begin{array}{ccc}15\\6\end{array}\right)(0.32)^{6}(1-0.32)^{15 - 6}$ + $\left(\begin{array}{ccc}15\\7\end{array}\right)(0.32)^{7}(1-0.32)^{15 - 7}$ + $\left(\begin{array}{ccc}15\\8\end{array}\right)(0.32)^{8}(1-0.32)^{15 - 8}$ + $\left(\begin{array}{ccc}15\\9\end{array}\right)(0.32)^{9}(1-0.32)^{15 - 9}$

= 0.213 + 0.1671 + 0.1011 + 0.0476 + 0.0174

= 0.5462

⇒P(5≤X≤9) = 0.5462

P(5<X<10) = + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

= $\left(\begin{array}{ccc}15\\6\end{array}\right)(0.32)^{6}(1-0.32)^{15 - 6}$ + $\left(\begin{array}{ccc}15\\7\end{array}\right)(0.32)^{7}(1-0.32)^{15 - 7}$ + $\left(\begin{array}{ccc}15\\8\end{array}\right)(0.32)^{8}(1-0.32)^{15 - 8}$ + $\left(\begin{array}{ccc}15\\9\end{array}\right)(0.32)^{9}(1-0.32)^{15 - 9}$

= 0.1671 + 0.1011 + 0.0476 + 0.0174

= 0.3332

⇒P(5<X<10) = 0.3332

P(X≥6) = 1 - P(X < 6)

= 1 - P(X ≤ 5)

= 1 - binomial(5, 15, 0.32, 1)

= 1 - 0.6607

= 0.3393

⇒P(X≥6) = 0.3393

P(X<5) = P(X≤4)

= binomial (4, 15, 0.32, 1)

= 0.4477

⇒P(X<5) = 0.4477

Mean = np

= 15(0.32)

= 4.8

⇒Mean = 4.8

Variance = np(1-p)

= 15(0.32)(1 - 0.32)

= 4.8(0.68)

= 3.264

⇒Variance = 3.264

Answer from: Quest

54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m 60 second 60 minutes

* * = 54000 meters/ hour

second 1 minute 1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m 1 km

* = 54 km / hours

hour 1000m

Answer from: Quest

Looks pretty good to me!

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