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.
- 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 \%$.