TWENTY POINTS

iogann1982 [59]

The Human Heart Beat 80 Times in a munite

3 0

4 years ago

A normally distributed set of numbers has a mean of 75 and a standard deviation of 7.97. What percentage of values lies between

lana [24]

Answer:

63% of the values lies between 70 and 85.

Step-by-step explanation:

We are given that a normally distributed set of numbers has a mean of 75 and a standard deviation of 7.97.

Let X = Set of numbers

The z-score probability distribution for is given by;

Z = $\frac{ X -\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean value = 75

$\sigma$ = standard deviation = 7.97

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percentage of values that lies between 70 and 85 is given by = P(70 < X < 85) = P(X < 85) - P(X $\leq$ 70)

P(X < 85) = P( $\frac{ X -\mu}{\sigma}$ < $\frac{ 85-75}{7.97}$ ) = P(Z < 1.25) = 0.89435 {using z table}

P(X $\leq$ 70) = P( $\frac{ X -\mu}{\sigma}$ $\leq$ $\frac{ 70-75}{7.97}$ ) = P(Z $\leq$ -0.63) = 1 - P(Z < 0.63)

= 1 - 0.73565 = 0.26435

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.25 and x = 0.63 in the z table which has an area of 0.89435 and 0.73565 respectively.

Therefore, P(70 < X < 85) = 0.89435 - 0.26435 = 0.63 or 63%

Hence, 63% of the values lies between 70 and 85.

6 0

4 years ago

Pleaseeeeeee I have to finish it in 15 minutes???

lys-0071 [83]

Answer:

it is too small i'm sorry

Step-by-step explanation:

6 0

4 years ago

Equivalent Expressions

Westkost [7]

Tell me what grade you in and maybe I.can message u on here if your in sixth grade or lower I can help

5 0

3 years ago

The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What

atroni [7]

Answer:

21.77% probability that a randomly selected teacher earns more than $525 a week

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 490, \sigma = 45$