Question

If the mean is 298 and standard deviation is 22 what is the probability that a random sample of 42 pregnancies has a mean gestation period of less than 290 days

Answer 1

Answer:

The probability is 0.9909.

Step-by-step explanation:

Test statistic (z) = (sample mean - population mean) ÷ (sd/√n)

sample mean = 290 days

population mean = 298 days

sd = 22 days

n = 42

z = (290 - 298) ÷ (22/√42) = -8 ÷ 3.395 = -2.36

The cumulative area of the test statistic is the probability that the mean gestation period is less than 290 days. The cumulative area is 0.9909. Therefore the probability is 0.9909.

Answer 2

Answer:

0.9909

Step-by-step explanation:

population mean = 298

standard deviation = 22

probability of 42 pregnancies having a mean of 290 = ?

N = 42

given mean = 290

solution:

The statistics ( Y ) = $\frac{(given mean - population mean)}{\frac{standard deviation}{\sqrt{N} } }$

                             = $\frac{290- 298}{\frac{22}{\sqrt{42} } }$ = $\frac{-8}{ \frac{22}{6.48} }$

                             = -8 / 3.3950

hence the probability of 42 pregnancies having a mean gestation period of 290 = 0.999

this is because the probability mean gestation period been calculated for is less than 290 days