Math / Data & Statistics

QuestionThe lengths of pregnancies in a small rural village is a normally distributed random variable X with a mean of 266 days and a standard deviation of 15 days.
What percentage of pregnancies last beyond 244 days? Enter answer as a decimal, round to 2 decimal places $P(X>244 \text { days })=$ $\square$ Question Help: Video

Studdy Solution

STEP 1

What is this asking? What proportion of pregnancies in this village last longer than 244 days, given that the lengths of pregnancies follow a normal distribution? Watch out! Don't forget to convert the number of days to a z-score before looking up the probability in the z-table!
Also, remember that the z-table gives the probability to the left of a z-score, so we'll need to do a little subtraction to get the probability to the right.

STEP 2

1. Calculate the z-score.
2. Find the probability associated with the z-score.
3. Calculate the probability of a pregnancy lasting longer than 244 days.

STEP 3

We start by calculating the z-score.
The z-score tells us how many standard deviations a particular value is away from the mean.
The formula for the z-score is:
$z = \frac{x - \mu}{\sigma}$ Where $x$ is the value we're interested in (in this case, $\textbf{244}$ days), $\mu$ is the mean ( $\textbf{266}$ days), and $\sigma$ is the standard deviation ( $\textbf{15}$ days).

STEP 4

Let's plug in our values:
$z = \frac{244 - 266}{15}$

STEP 5

Simplifying the numerator:
$z = \frac{-22}{15}$

STEP 6

Calculating the division:
$z \approx -1.47$ So, 244 days is approximately 1.47 standard deviations below the mean pregnancy length.

STEP 7

Now, we need to find the probability associated with a z-score of $\textbf{-1.47}$ .
We can look this up in a z-table (or use a calculator).
The z-table tells us the probability of a value being less than a given z-score.

STEP 8

Looking up $z = -1.47$ in the z-table, we find a probability of approximately $\textbf{0.0708}$ .
This means that there's a 7.08% chance of a pregnancy lasting less than 244 days.

STEP 9

Since we want the probability of a pregnancy lasting longer than 244 days, we need to subtract the probability we just found from 1.
Remember, the total probability under the normal distribution curve is always 1.

STEP 10

Calculation:
$P(X > 244) = 1 - P(X < 244)$ $P(X > 244) = 1 - 0.0708$ $P(X > 244) \approx 0.9292$

STEP 11

Approximately $\textbf{0.93}$ or 93% of pregnancies in this village last longer than 244 days.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the z-score.
2. Find the probability associated with the z-score.
3. Calculate the probability of a pregnancy lasting longer than 244 days.

STEP 3

STEP 4

Let's plug in our values:
$z = \frac{244 - 266}{15}$

STEP 5

Simplifying the numerator:
$z = \frac{-22}{15}$

STEP 6

Calculating the division:
$z \approx -1.47$ So, 244 days is approximately 1.47 standard deviations below the mean pregnancy length.

STEP 7

Now, we need to find the probability associated with a z-score of $\textbf{-1.47}$ .
We can look this up in a z-table (or use a calculator).
The z-table tells us the probability of a value being less than a given z-score.

STEP 8

Looking up $z = -1.47$ in the z-table, we find a probability of approximately $\textbf{0.0708}$ .
This means that there's a 7.08% chance of a pregnancy lasting less than 244 days.

STEP 9

Since we want the probability of a pregnancy lasting longer than 244 days, we need to subtract the probability we just found from 1.
Remember, the total probability under the normal distribution curve is always 1.

STEP 10

Calculation:
$P(X > 244) = 1 - P(X < 244)$ $P(X > 244) = 1 - 0.0708$ $P(X > 244) \approx 0.9292$

STEP 11

Approximately $\textbf{0.93}$ or 93% of pregnancies in this village last longer than 244 days.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the *z*-score.2. Find the probability associated with the *z*-score.3. Calculate the probability of a pregnancy lasting longer than 244 days.

STEP 3

STEP 4

Let's **plug in** our values:z=244−26615 z = \frac{244 - 266}{15} z=15244−266​

STEP 5

**Simplifying** the numerator:z=−2215 z = \frac{-22}{15} z=15−22​

STEP 6

**Calculating** the division:z≈−1.47 z \approx -1.47 z≈−1.47So, 244 days is approximately **1.47 standard deviations below** the mean pregnancy length.

STEP 7

Now, we need to find the probability associated with a *z*-score of -1.47\textbf{-1.47}-1.47.We can look this up in a *z*-table (or use a calculator).The *z*-table tells us the probability of a value being *less than* a given *z*-score.

STEP 8

Looking up z=−1.47z = -1.47z=−1.47 in the *z*-table, we find a probability of approximately 0.0708\textbf{0.0708}0.0708.This means that there's a 7.08% chance of a pregnancy lasting *less than* 244 days.

STEP 9

Since we want the probability of a pregnancy lasting *longer* than 244 days, we need to subtract the probability we just found from 1.Remember, the total probability under the normal distribution curve is always 1.

STEP 10

**Calculation:**P(X>244)=1−P(X<244) P(X > 244) = 1 - P(X < 244) P(X>244)=1−P(X<244)P(X>244)=1−0.0708 P(X > 244) = 1 - 0.0708 P(X>244)=1−0.0708P(X>244)≈0.9292 P(X > 244) \approx 0.9292 P(X>244)≈0.9292

STEP 11

Approximately 0.93\textbf{0.93}0.93 or **93%** of pregnancies in this village last longer than 244 days.

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1. Calculate the z-score.
2. Find the probability associated with the z-score.
3. Calculate the probability of a pregnancy lasting longer than 244 days.

Let's plug in our values:
$z = \frac{244 - 266}{15}$

Simplifying the numerator:
$z = \frac{-22}{15}$

Calculating the division:
$z \approx -1.47$ So, 244 days is approximately 1.47 standard deviations below the mean pregnancy length.

Now, we need to find the probability associated with a z-score of $\textbf{-1.47}$ .
We can look this up in a z-table (or use a calculator).
The z-table tells us the probability of a value being less than a given z-score.

Looking up $z = -1.47$ in the z-table, we find a probability of approximately $\textbf{0.0708}$ .
This means that there's a 7.08% chance of a pregnancy lasting less than 244 days.

Since we want the probability of a pregnancy lasting longer than 244 days, we need to subtract the probability we just found from 1.
Remember, the total probability under the normal distribution curve is always 1.

Calculation:
$P(X > 244) = 1 - P(X < 244)$ $P(X > 244) = 1 - 0.0708$ $P(X > 244) \approx 0.9292$

Approximately $\textbf{0.93}$ or 93% of pregnancies in this village last longer than 244 days.