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 p**opulation 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.**

**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 \%$.