4) An economist is interest in studying the incomes of consumers in a particular country. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in a mean income of $15,000. What total sample size would the economist need to use for a 95% confidence interval if the width of the interval should NOT be more than $100

Population standard deviation б =1000

95% confidence interval width should not be more than $100.

Hence, (x+E) - (x-E) ≤ 100

2E ≤ 100

E ≤ 100/2

E ≤ 50

Level of significance is ∝ = 0.05 . Using the normal area table values at 0.05, the critical value is Z(∝/2) = 1.96

Computation of the sample size required.

n = [ (Z(∝/2) * б) / E]^2

n = [1.96  * 1000 / 50]^2

n = 39.2^2

n = 1536.64

n = 1537

Hence, the economist needed a sample size of 1537 for a 95% confidence interval if the width of the interval should not be more than $100.

the expected value of the random variable : 0, 1 and 2

step-by-step explanation:

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752