Question

The lengths of human pregnancies are normally distributed with a mean of $268\text{days}$ and a standard deviation of $15\text{days}$.

What is the probability that a pregnancy lasts at least $300\text{days}$?

• A. $0.4834$
• B. $0.0166$
• C. $0.0332$
• D. $0.9834$
• E. $0.0179$

Answer 1

The correct option is B

$0.0166$

Determine the probability that pregnancy lasts at least $300\text{days}$:

The explanation for the correct option:

For a normally distributed random variable $x$ with mean $\mu$ and standard deviation $\sigma$, the probability $P(z<c)$ that the $z$-score $z=\frac{x-\mu }{\sigma }$ is less than a particular number $c$ can be looked up from a standard normal distribution $z$-score table.

It is given that length of human pregnancies is a random variable distributed normally with a mean of $268\text{days}$ and a standard deviation of $15\text{days}$.

It is required to find the probability that a pregnancy lasts at least $300\text{days}$.

Find the $z$-score $z=\frac{x-\mu }{\sigma }$ for $x=300$, $\mu =268$, and $\sigma =15$.

$\begin{array}{rcl}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{300-268}{15}\\ & =& \frac{32}{15}\\ & \approx & 2.1333\end{array}$

Thus the probability $P(x\ge 300)$ is equal to $P(z\ge 2.1333)$.

Note that the event $z\ge 2.1333$ is complement of the event $z<2.1333$.

Since, probabilities of complementary events add up to $1$, write $P(z\ge 2.1333)=1-P(z<2.1333)$.

Look up $P(z<2.1333)$ from a standard normal distribution $z$-score table to get $P(z<2.1333)=0.98355$.

Substitute this value in the equation $P(z\ge 2.1333)=1-P(z<2.1333)$.

$\begin{array}{rcl}P(z& \ge & 2.1333)=1-P(z<2.1333)\\ & =& 1-0.98355\\ & =& 0.01645\end{array}$

Thus, the probability that pregnancy lasts at least $\mathbf{300}$ days is approximately $\begin{array}{l}0.01645\end{array}$.

Check the options for correctness.

The options $A$, $C$, $D$, and $E$ are $0.4834$, $0.0332$, $0.9834$, $0.0179$ respectively,

All these options differ significantly from the probability calculated above $\begin{array}{l}0.01645\end{array}$.

Thus, options $A$, $C$, $D$ and $E$ are incorrect.

The Option $B$ $0.0166$ is sufficiently close to the probability calculated above $\begin{array}{l}0.01645\end{array}$.

Thus, Option $B$ is the most suitable answer.

Hence, options $A$, $C$, $D$, and $E$ are incorrect. Option $\left(B\right)$ $\mathbf{0}\mathbf{.}\mathbf{0166}$ is the correct answer.