QuestionFind the standard deviation of human pregnancy lengths given a mean of 267 days and $95\%$ lasting between 245 and 289 days. What percent last at least 285 days? Standard deviation: $\square$ days.

Studdy Solution

STEP 1

Assumptions1. The mean length of a human pregnancy is267 days. .95% of all human pregnancies last between245 and289 days.
3. The distribution of lengths of human pregnancies follows a normal distribution.

STEP 2

The95% confidence interval for a normal distribution corresponds to approximately2 standard deviations from the mean. This means that95% of the data falls within2 standard deviations of the mean. We can use this information to calculate the standard deviation.
$Standard\, deviation = (Upper\, limit - Lower\, limit) /4$

STEP 3

Plug in the given values for the upper limit and lower limit to calculate the standard deviation.
$Standard\, deviation = (289 -245) /$

STEP 4

Calculate the standard deviation.
$Standard\, deviation = (289 -245) /4 =11$ The standard deviation of the distribution of lengths of human pregnancies is11 days.

STEP 5

To find the percentage of pregnancies that last at least285 days, we need to find the z-score for285 days. The z-score is a measure of how many standard deviations an element is from the mean.
$Z = (X - \mu) / \sigma$

STEP 6

Plug in the given values for X (285 days), the mean (267 days), and the standard deviation (11 days) to calculate the z-score.
$Z = (285 -267) /11$

STEP 7

Calculate the z-score.
$Z = (285 -267) /11 =1.64$

STEP 8

The z-score of1.64 corresponds to a percentile of approximately95% in the standard normal distribution. This means that approximately95% of pregnancies last less than285 days. Therefore, the percentage of pregnancies that last at least285 days is100% -95% =5%.
The standard deviation of the distribution of lengths of human pregnancies is11 days and5% of human pregnancies are expected to last at least285 days.

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QuestionFind the standard deviation of human pregnancy lengths given a mean of 267 days and $95\%$ lasting between 245 and 289 days. What percent last at least 285 days? Standard deviation: $\square$ days.

Studdy Solution

STEP 1

Assumptions1. The mean length of a human pregnancy is267 days. .95% of all human pregnancies last between245 and289 days.
3. The distribution of lengths of human pregnancies follows a normal distribution.

STEP 2

STEP 3

Plug in the given values for the upper limit and lower limit to calculate the standard deviation.
$Standard\, deviation = (289 -245) /$

STEP 4

Calculate the standard deviation.
$Standard\, deviation = (289 -245) /4 =11$ The standard deviation of the distribution of lengths of human pregnancies is11 days.

STEP 5

To find the percentage of pregnancies that last at least285 days, we need to find the z-score for285 days. The z-score is a measure of how many standard deviations an element is from the mean.
$Z = (X - \mu) / \sigma$

STEP 6

Plug in the given values for X (285 days), the mean (267 days), and the standard deviation (11 days) to calculate the z-score.
$Z = (285 -267) /11$

STEP 7

Calculate the z-score.
$Z = (285 -267) /11 =1.64$

STEP 8

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QuestionFind the standard deviation of human pregnancy lengths given a mean of 267 days and 95%95\%95% lasting between 245 and 289 days. What percent last at least 285 days? Standard deviation: □\square□ days.

Studdy Solution

STEP 1

Assumptions1. The mean length of a human pregnancy is267 days. .95% of all human pregnancies last between245 and289 days.3. The distribution of lengths of human pregnancies follows a normal distribution.

STEP 2

STEP 3

Plug in the given values for the upper limit and lower limit to calculate the standard deviation.Standard deviation=(289−245)/Standard\, deviation = (289 -245) /Standarddeviation=(289−245)/

STEP 4

Calculate the standard deviation.Standard deviation=(289−245)/4=11Standard\, deviation = (289 -245) /4 =11Standarddeviation=(289−245)/4=11The standard deviation of the distribution of lengths of human pregnancies is11 days.

STEP 5

To find the percentage of pregnancies that last at least285 days, we need to find the z-score for285 days. The z-score is a measure of how many standard deviations an element is from the mean.Z=(X−μ)/σZ = (X - \mu) / \sigmaZ=(X−μ)/σ

STEP 6

Plug in the given values for X (285 days), the mean (267 days), and the standard deviation (11 days) to calculate the z-score.Z=(285−267)/11Z = (285 -267) /11Z=(285−267)/11

STEP 7

Calculate the z-score.Z=(285−267)/11=1.64Z = (285 -267) /11 =1.64Z=(285−267)/11=1.64

STEP 8

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the standard deviation of human pregnancy lengths given a mean of 267 days and $95\%$ lasting between 245 and 289 days. What percent last at least 285 days? Standard deviation: $\square$ days.

Assumptions1. The mean length of a human pregnancy is267 days. .95% of all human pregnancies last between245 and289 days.
3. The distribution of lengths of human pregnancies follows a normal distribution.

Plug in the given values for the upper limit and lower limit to calculate the standard deviation.
$Standard\, deviation = (289 -245) /$

Calculate the standard deviation.
$Standard\, deviation = (289 -245) /4 =11$ The standard deviation of the distribution of lengths of human pregnancies is11 days.

To find the percentage of pregnancies that last at least285 days, we need to find the z-score for285 days. The z-score is a measure of how many standard deviations an element is from the mean.
$Z = (X - \mu) / \sigma$

Plug in the given values for X (285 days), the mean (267 days), and the standard deviation (11 days) to calculate the z-score.
$Z = (285 -267) /11$

Calculate the z-score.
$Z = (285 -267) /11 =1.64$