Please help... I hate algebra

The original pulse rates are measure with units of "beats per minute". what are the units of the corresponding z scores? choo

posledela

the complete question is

Pulse rates of women are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute. The original pulse rates are measure with units of "beats per minute". What are the units of the corresponding z scores? Choose the correct choice below.

A. The z scores are measured with units of "beats per minute."

B. The z scores are measured with units of "minutes per beat."

C. The z scores are measured with units of "beats."

D. The z scores are numbers without units of measurement.

we know that

A z-score is the number of standard deviations from the mean a data point is. But more technically it’s a measure of how many standard deviations below or above the population mean a raw score is

The z scores are numbers without units of measurement

therefore

the answer is the option D

8 0

3 years ago

suppose that the distribution for total amounts spent by students vacationing for a week in florida is normally distributed with

pentagon [3]

The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is 0.8132.

Let x be the total amount spent by students.

x follows normal distribution with mean μ = 650,

standard deviation δ = 120

We take a simple random sample of size n = 10

We are asked to find average spending of 10 students ( $x$ ) is between 600 and 700

We have to find P( 600 <= $x$ <= 700)

According to the sampling distribution of the sample mean $x$ , it follows an approximately normal distribution with mean μ{ $x$ } = μ and standard deviation δ{ $x$ } = {δ}/{√n}

Therefore here mean of $x$ ,( μ{ $x$ } ) = 650 and

standard deviation of $x$ , δ{ $x$ } = {120}/{√10} = 37.9473

The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is,

Let Z= x - μ / δ

Z₁ = 600 - 650 / 37.9473 = -1.32 similarly

Z₂ = 700 - 650 / 37.9473 = 1.32

From standard normal distribution table, P( -1.32 < $x$ < 1.32) = 0.8132

The probability that the SRS of 10 students will spend an average of between 600 and 700 dollars is 0.8132

To learn more about Normal distribution click here:

brainly.com/question/15103234

#SPJ4

3 0

1 year ago

The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance

lisabon 2012 [21]

Answer:

The 95% confidence interval for the variance in the pounds of impurities would be $3.829 \leq \sigma^2 \leq 14.121$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =6.62$ represent the sample variance

s=2.573 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical values.