Question

A Food Marketing Institute found that 27% of households spend more than $125 a week on groceries. Assume the population proportion is 0.27 and a simple random sample of 193 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.29? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

dim: To find probability that the sample proportion of households spending more than $195 than $125 a weekis deop than 0:29. Given : population propottion =p=0.27. sample size = n = 193 Alolution : let ſ be the pample proportion. we know, 21-22 Î NMCPF

3–2 ر نہ Ninfo 0, Pli-B NMCO, 1). P-p PUR Do bere we have to actually find probability that sample pro poztion is less than 0.89. Pel@ < 0:29). pcem < 0.29) Plip pcb- Plo-e < 0.29 0.29 - 0.27 NCO, 1) < 0:27 (1-0-272 193 NCO,L) < If 0-02 0.0319 6

-PINCOIL) < 0.6258). = 1- PINIO,1) > 0.6258 Normalip a distribution Bymmetric - 1-PC N10,1) > 0.63) = 1- 0.26435 0.7359 = pzom otatístical table. PC <0.29) = Answers The proba bility that the sample proportion of households epending more than $ 125 a week being less than 0.29 is 0.7357