Ask question

Ask question

All categories

2 years ago

8

Assume that females have pulse rates that are normally distributed with a mean of mu equals 73.0 beats per minute and a standard

deviation of sigma equals 12.5 beats per minute. Complete parts (a) through (c) below. a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 69 beats per minute and 77 beats per minute. The probability is nothing. (Round to four decimal places as needed.)

1 answer:

slava [35]2 years ago

4 0

Answer: 0.2510

Step-by-step explanation:

Given : Mean = $\mu=\text{73.0 beats per minute}$

Standard deviation : $\sigma=\text{12.5 beats per minute}$

Sample size : n=1

We assume that females have pulse rates that are normally distributed.

Then , the formula to calculate the z-score is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x =69

$z=\dfrac{69-73}{\dfrac{12.5}{\sqrt{1}}}=-0.32$

For x= 77

$z=\dfrac{77-73}{\dfrac{12.5}{\sqrt{1}}}=0.32$

The p-value = $P(-0.32$

$=P(z$

Hence, the probability that her pulse rate is between 69 beats per minute and 77 beats per minute =0.2510

You might be interested in

A cube with side length 4p is stacked on another cube with side length 2q^2. What is the total volume of the cubes in factored f

AnnZ [28]

Volume of cube=side³
ok, so you need to know the difference or sum of cubes
a³+b³=(a+b)(x²-xy+y²)
so

(4p)³+(2q²)³=
(4p+2q²)((4p)²-(4p)(2q²)+(2q²)²)=
(4p+2q²)(16p²-8pq²+4q⁴)

3rd option

5 0

1 year ago

A local project being analyzed by PERT has 42 activities, 13 of which are on the critical path. If the estimated time along the

ryzh [129]

Answer:

the probability that the project will be completed in 95 days or less, P(x ≤ 95) = 0.023

Step-by-step explanation:

This is a normal probability distribution question.

We'll need to standardize the 95 days to solve this.

The standardized score is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ

x = 95 days

xbar = mean = 105 days

σ = standard deviation = √(variance) = √25 = 5

z = (95 - 105)/5 = - 2

To determine the probability that the project will be completed in 95 days or less, P(x ≤ 95) = P(z ≤ (-2))

We'll use data from the normal probability table for these probabilities

P(x ≤ 95) = P(z ≤ (-2)) = 0.02275 = 0.023

5 0

2 years ago

290 pennies are equal to how many dimes

ExtremeBDS [4]

29 dimes
............

6 0

2 years ago

Read 2 more answers

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

1 year ago

Write each fraction as a sum of unit fractions 7/10

seropon [69]

Answer:

3/10+4/10=7/10

Step-by-step explanation:

8 0

2 years ago