Assume that the duration of human pregnancies can be described by a Normal model with mean 266 days and standard deviation 16 days. a) What percentage of pregnancies should last between 270 and 280 days? b) At least how many days should the longest 25% of all pregnancies last? c) Suppose a certain obstetrician is currently providing prenatal care to 60 pregnant women. Let y̅ represent the mean length of their pregnancies. According to the Central Limit Theorem, what’s the distribution of this sample mean, y̅? Specify the model, mean, and standard deviation. d) What’s the probability that the mean duration of these patient’s pregnancies will be less than 260 days?

Assume That The Duration Of Human Pregnancy

This article aims to find the z-score values for the different conditions with $ \mu $ and $\sigma $. The article uses the concept of z-score and z-table. Simply put, the z-score (also called the standard score) gives you an idea of how far a data point is from the mean. But more technically, it’s a measure of how many standard deviations below or above the population mean the raw score is. The formula for the z-score is given as:

\[z = \dfrac { x – \mu }{ \sigma } \]

Expert Answer

Part (a)

The mean and standard deviation is given as:

\[\mu = 266 \]

\[ \sigma =16 \]

\[P( 270 \leq X \leq 280 ) = P (\dfrac {270 – 266} {16} \leq z \leq \dfrac {280 – 266 }{16})  = P(0.25 \leq z \leq 0.88) \]

\[P (0.25 \leq z \leq 0.88) = P(z \leq 0.88) – P(z \leq 0.25) \]

\[=0.8106-0.5987 \]

\[ = 0.2119\]

Percentage of pregnancies that should last between $270$ and $280$ days will therefore be $21.1\% $

Part (b)

\[P ( Z  \geq z ) = 0.25 \]

By using $ z-table $

\[ z = 0.675 \]

\[ \dfrac { x – 266 }{ 16 } = 0.675 \]

\[ x = 276.8 \]

So the longest $ 25\% $ of all pregnancies should last at least $ 277 $ days.

Part (c)

The shape of the sample distribution model for the mean pregnancy will be a normal distribution.

\[ \mu = 266 \]

\[ \sigma = \dfrac { 16 }{ \sqrt 60 } = 2.06 \]

Part (d)

\[P (X \leq 260 ) = P (z \leq \dfrac { 260 – 266 } { 2.06 } ) = P( z \leq -2.914) = 0.00187 \]

So the probability that the average length of pregnancy will be less than $260$ days is $0.00187$.

Numerical Result

(a)

Percentage of pregnancies that last between $270$ and $280$ days will therefore be $21.1\%$

(b)

The longest $25\%$ of all pregnancies should last at least $277$ days.

(c)

The shape of the sample distribution model for the mean pregnancy will be a normal distribution with mean $ \mu = 266 $ and standard deviation $\sigma =2.06 $.

(d)

The probability that the average length of pregnancy will be less than $260$ days is $0.00187$.

Example

Assume that a standard model can describe the duration of human pregnancies with a mean of $270$ days and a standard deviation of $18$ days.

  1. a) What is the percentage of pregnancies that last between $280$ and $285$ days?

Solution

Part (a)

The mean and standard deviation is given as:

\[\mu = 270 \]

\[ \sigma = 18 \]

\[P( 280 \leq X \leq 285 ) = P (\dfrac {280-270}{18} \leq z \leq \dfrac {285-270}{18} )  = P(0.55 \leq z \leq 0.833) \]

\[P (0.55 \leq z \leq 0.833) = P (z \leq 0.833) – P (z \leq 0.55) \]

\[= 0.966 – 0.126 \]

\[ = 0.84 \]

Percentage of pregnancies that should last between $280$ and $285$ days will therefore be $ 84 \%$.

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