Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 269 days and a standard deviation of 13 days. A distribution of values is normal with a mean of 269 and a standard deviation of 13.

What percentage of pregnancies last beyond 276 days?
P(X > 276 days) = ____%

Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Add Work

Answer 1

Solution :

Given that ,

mean = $\mu$ = 269

standard deviation = $\sigma$ = 13

P(x > 276) = 1 - P(x < 276)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (276 - 269) /13 )

= 1 - P(z < 0.538)

= 1 - 0.7047

0.2953

P(x > 276) = 29.5