A random sample of 225 individuals working in a large city indicated that 45 are dissatisfied with their working conditions. Based upon this, compute a 90% confidence interval for the proportion of all individuals in this city who are dissatisfied with their working conditions. Then find the lower limit and upper limit of the 90% confidence interval.

CI 90 %  =  [  0,165  ;  0,235 ]

Lower limit     0,165

upper limit      0,235

Step-by-step explanation:

Sample information:

sample size    n  =  225

number of dissatisfied  individuals   x  =  45

p = 45/225

p = 0,2    and  q =  1 -  p   q  =  1  -  0,2   q  =  0,8

p*n = 0,2*225   =  45    and   q*n   =  0,8*225  =  180

p*n  and q*n  big enough to use the approximation of binomial distribution to normal distribution

90 % of Confidence Interval   then a significance level is  α  = 10%

α  =  0,1 in z table we get z(c) for that significance level

z(c) = 1,28

CI 90 %   =  [  p  ±  z(c)*√(p*q)/n ]

z(c) * √(p*q)/n    =  1,28 *  √ ( 0,2*0,8)/225

z(c) * √(p*q)/n    =  1,28 * 0,027

z(c) * √(p*q)/n    =  0,035

CI 90 %  =  [  0,2  ±  0,035 ]

CI 90 %  =  [  0,165  ;  0,235 ]

Lower limit     0,165

upper limit      0,235

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.

your answer   is c

step-by-step explanation: