Question

Please help me with this

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 38 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 38 weeks and that the population standard deviation is 6 weeks. Suppose you would like to select a random sample of 35 unemployed individuals for a follow-up study Find the probability that a single randomly selected value is less than 39. P(X<39) 0.3949 Find the probability that a sample of size n = 35 is randomly selected with a mean less than 39. PM < 39) = | 0.0220 |丼 Enter your answer as numbern accurac to 4 decimal places

Answer 1

We know from central limit theorem, if the size of sample is large (30 or more) , then the sampling distribution of sample mean is approximately normal i.e.

$\fn_cm \bar{x}\sim N\left (\mu ,\frac{\sigma ^{2}}{n} \right )$   

and $\fn_cm x\sim N\left (\mu ,\sigma ^{2} \right )$

We have  $\fn_cm \mu =38$, $\fn_cm \sigma =6$ and $\fn_cm n=35$

• The probability that a single randomly selected value is less than 39

$\fn_cm P(X< 39) = P\left ( \frac{X- \mu }{\sigma } < \frac{39 - \mu }{\sigma }\right )$

$\fn_cm = P\left ( Z < \frac{39 - 38}{6 }\right )$

$\fn_cm = P\left ( Z <0.1667)$

  $\fn_cm = 0.566$ (Using the standard normal table)

• The probability that a sample of size n = 35 is randomly selected with a mean less than 39.

$\fn_cm P(M< 39) = P\left ( \frac{M- \mu }{\sigma/\sqrt{n} } < \frac{39 - \mu }{\sigma/\sqrt{n} }\right )$

  $\fn_cm = P\left ( Z < \frac{39 - 38 }{6/\sqrt{35} }\right )$

$\fn_cm = P\left ( Z < 0.9859)$

$\fn_cm =0.8389$

Now, we have  $\fn_cm \mu =35$, $\fn_cm \sigma =3.1$ and $\fn_cm n=87$

• The probability that a single randomly selected value is greater than 35.3

$\fn_cm P(X> 35.3) = P\left ( \frac{X- \mu }{\sigma } < \frac{35.3- \mu }{\sigma }\right )$

$\fn_cm = P\left ( Z > \frac{35.3 - 35}{3.1 }\right )$

$\fn_cm = P\left ( Z > 0.09677)$

$\fn_cm = 0.4602$ (using normal table)

0.16 0.14 0.12 0.10 0.06 0.04 0.02 0.00 0.02 34 36 38 40 42 46 24 26 28 30 32

  

• The probability that a sample of size n = 87 is randomly selected with a mean greater than 35.3

$\fn_cm P(M> 35.3) = P\left ( \frac{M- \mu }{\sigma/\sqrt{n} } > \frac{35.3 - \mu }{\sigma/\sqrt{n} }\right )$

  $\fn_cm = P\left ( Z > \frac{35.3 - 35 }{3.1/\sqrt{87} }\right )$

$\fn_cm = P\left ( Z > 0.9027)$

$\fn_cm =0.1841$