Question

About 6% of employed adults in the United States held multiple jobs. A random sample of 69 employed adults is chosen. Use th TI-84 Plus calculator as needed. Part 1 of 5 (a) Is it appropriate to use the normal approximation to find the probability that less than 6.6% of the individuals in the sample hold multiple jobs? If so, find the probability. If not, explain why not. It is appropriate to use the normal curve, since np - 4.14 < 10. Part 2 of 5 (b) A new sample of 339 employed adults is chosen. Find the probability that less than 6.6% of the individuals in this sample hold multiple jobs. Round the answer to at least four decimal places. The probability that less than 6.6% of the individuals in this sample hold multiple jobs is 0.6791

Answer 1

(a) It is not appropriate to use Normal curve , since np = 69*0.06= 4.14 < 10

(b) As n = 339

np = 339*0.06 = 20.34 > 10 , n(1-p) = 339*(1-0.06) = 318.66> 10

then we can use normal approximation to binomial

As we know that sampling distribution of sample proportion ($\hat{p}$ ) follow Normal  

with mean, E( $\hat{p}$ ) = p = 0.06

and standard error = $\sqrt{\hat{p}(1-\hat{p})/n}= \sqrt{0.06*0.94/339}=0.0129$

To find

$P(\hat{p}< 0.066)$

As $\hat{p}\sim N(0.06,0.0129)$

$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}\sim N(0,1)$

$P(\hat{p}< 0.066)$

$=P(z<\frac{0.066-0.06}{0.0129})$

= P( z < 0.47)

= 0.6808

Answer : Probability = 0.6808

(c)

$P(\hat{p}> 0.061)$

$=P(z>\frac{0.061-0.06}{0.0129})$

= P( z > 0.08)

= 0.4681

Answer : Probability = 0.4681

Note : Using TI 84 calculator , P( z < 0.47)

Press 2nd key , then VARS , chose normal cdf ,

lower limit = -99

upper limit = 0.47

mean =0 , stanadard deviation = 1

that is enter normalcdf( -99, 0.47, 0,1) we get P( z< 0.47) = 0.6808

For P( z > 0.08 ) ,we first find P( z < 0.08)

that is enter normalcdf( -99, 0.08, 0,1) we get P( z< 0.08 ) = 0.5319

p( z > 0.08) = 1- 0.5319 = 0.4681