Question

Theorem ork- The Central Limit A leading magazine(like Barron's) reported at one time that the average number of weeks an individual is unemployed is 34 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 34 weeks and that the population standard deviation is 3.5 weeks. Suppose you would like to select a random sample of 55 unemployed individuals for a follow-up study Find the probability that a single randomly selected value is greater than 34.3. Px>34.3)(Enter your answers as numbers accurate to 4 decimal places.) Find the probability that a sample of size n # 55 is randomly selected with a mean greater than 34.3. PM 343)Enter your answers as numbers accurate to 4 decimal places.) Points possible:1 Unlimited attempts.

Answer 1

1)

Here, $X\sim N(34,3.5^{2})$

$P(X>34.3)=P(Z>\frac{34.3-34}{3.5})=P(Z>0.09)=1-P(Z<0.09)=1-0.5359=0.4641$

Also, $M\sim N(\mu,\frac{\sigma^{2}}{n})\sim N(34,\frac{3.5^{2}}{55})$

$P(M>34.3)=P(M>\frac{34.3-34}{\sqrt{\frac{3.5^{2}}{55}}})=P(Z>0.64)=1-P(Z<0.64)=1-0.7389=0.2611$

2)

Here, $X\sim N(20.4,10^{2})$

$P(11.7<X<21.6)=P(\frac{11.7-20.4}{10}<Z<\frac{21.6-20.4}{10})$$=P(-0.87<Z<0.12)=P(Z<0.12)-P(Z<-0.87)$

$=P(Z<0.12)-P(Z>0.87)=P(Z<0.12)-[1-P(Z<0.87)]$

$=0.5478-[1-0.8078]=0.3556$

Also, $M\sim N(\mu,\frac{\sigma^{2}}{n})\sim N(20.4,\frac{10^{2}}{11})$

$P(11.7<M<21.6)=P(\frac{11.7-20.4}{\sqrt{\frac{10^{2}}{11}}}<Z<\frac{21.6-20.4}{\sqrt{\frac{10^{2}}{11}}})$

$=P(-2.89<Z<0.40)=P(Z<0.40)-P(Z<-2.89)$

$=P(Z<0.40)-P(Z>2.89)=P(Z<0.40)-[1-P(Z<2.89)]$

$=0.6554-[1-0.9981]=0.6535$