Respuesta :
Answer:
Probability that in a random sample of six cities, the sample mean would be more than 40 is 0.3372.
Step-by-step explanation:
We are given that the percent of births to mothers with less than a college education in all of the most populated cities in the U.S. has an average of 39.3 with a standard deviation of 4.1.
Assuming the data follows distribution. Also, a random sample of six cities is selected.
Firstly, Let [tex]\bar X[/tex] = sample mean of six cities
The z score probability distribution for sample mean is given by;
Z = [tex]\frac{ \bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = average percent of births = 39.3
[tex]\sigma[/tex] = standard deviation = 4.1
n = sample of cities = 6
Probability that in a random sample of six cities, the sample mean would be more than 40 is given by = P([tex]\bar X[/tex] > 40)
P([tex]\bar X[/tex] > 40) = P( [tex]\frac{ \bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{ 40-39.3}{\frac{4.1}{\sqrt{6} } }[/tex] ) = P(Z > 0.42) = 1 - P(Z [tex]\leq[/tex] 0.42)
= 1 - 0.66276 = 0.3372
Therefore, probability that the sample mean would be more than 40 is 0.3372.
