Help me pls I need help with this test

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a

spin [16.1K]

Answer:

$Mean = 68.9$

$s^2 =18.1$ --- Variance

Step-by-step explanation:

Given

$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

$x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5$

For 59 - 66

$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

For 75 - 82

$x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5$

For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

So, the table becomes:

$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

$Mean = \frac{2547.5}{37}$

$Mean = 68.9$ -- approximated

Solving (b): The sample variance:

This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

5 0

2 years ago

A pet store has 9dog leashes. It has 8 Fewer dog leashes than the dog collars. How many dog collars does the store have?

Katen [24]

Answer:

Since you know that the pet store has 9 dog leashed, and we know that they have 8 fewer dog leashes than dog collars we can just add 8 to 9 to get 17 dog collars

4 0

2 years ago

Read 2 more answers

How many solutions does the equation 6y - 3y -7 = -2 +3 have?

kenny6666 [7]

Answer:

6y - 3y - 7 = -2 +3

Simplify both sides:

3y -7 = 1

Add 7 to both sides:

3y = 8

Divide both sides by 3:

y = 8/3 = 2 2/3

There is only one solution.

8 0

3 years ago

A real estate agent wants to know how many owners of homes worth over $1,000,000 might be considering putting their home on the

JulsSmile [24]

Answer:

NO

Step-by-step explanation:

The changeability of a sampling distribution is measured by its variance or its standard deviation. The changeability of a sampling distribution depends on three factors:

N: The number of observations in the population.
n: The number of observations in the sample.
The way that the random sample is chosen.

We know the following about the sampling distribution of the mean. The mean of the sampling distribution (μ_x) is equal to the mean of the population (μ). And the standard error of the sampling distribution (σ_x) is determined by the standard deviation of the population (σ), the population size (N), and the sample size (n). That is

μ_x=p

σ_x== [ σ / sqrt(n) ] * sqrt[ (N - n ) / (N - 1) ]

In the standard error formula, the factor sqrt[ (N - n ) / (N - 1) ] is called the finite population correction. When the population size is very large relative to the sample size, the finite population correction is approximately equal to one; and the standard error formula can be approximated by:

σ_x = σ / sqrt(n).

6 0

2 years ago

Can you help me solve this word problem?

WITCHER [35]

I hope this helps you

if 18 gallons for 420 miles

? gallons for 357 miles

?.420=18.357

?=15,3

3 0

2 years ago

Read 2 more answers