7x+10 on a table …. Linear function

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0

2 years ago

The position function for a free-falling object is s(t) = -16t^2 + vt +5. A ball is thrown upward from the top of a 400-foot bui

mixer [17]

Answer:

6.21 seconds

Step-by-step explanation:

The position function of a free-falling object is,

$s(t) = -16t^2+v_0t+s_0$

Where,

$s_0$ = initial height of the object,

$v_0$ = initial velocity of the object,

Here,

$s_0=400\text{ feet}$

$v_0=35\text{ feet per sec}$

Hence, the height of the ball after t seconds,

$s(t)=-16t^2+35t+400$

When s(t) = 0,

$\implies -16t^2+35t+400=0$

$t=\frac{-35\pm \sqrt{35^2-4\times -16\times 400}}{2\times -16}$

$t=\frac{-35\pm \sqrt{1225+25600}}{-32}$

$\implies t\approx -4.02\text{ or }t\approx 6.21$

∵ Time cannot be negative,

Hence, after 6.21 seconds the ball will reach the ground.

4 0

3 years ago

4. Find the value of 8 + 12 + 4 + 6 x 1 + 2. Guys I am doing a test and I need help!!

skelet666 [1.2K]

32 is the correct answer I believe if using PEMDAS method

6 0

2 years ago

Which of the following statements is false regarding randomization?

MrRissso [65]

Answer:

II) randomization helps avoid blas

Step-by-step explanation:

3 0

3 years ago

Please help me graph #9.

Nata [24]

Hint to start out the problem:

eg. For f(4), use the first expression, (x-3)^2 because 4 is greater than 1.

4 0

3 years ago

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