Question

Using Normal Distribution to Analyze Data Martin suspects that people carry less cash with them than they used to because of credit cards and other cashless payment systems. He decides to work on inventing a new system for people to use in place of cash that would be more user friendly and safe. Martin conducted a survey to help convince investors to back the development of his new cashless system. One afternoon at the local mall, Martin surveyed a random sample of 95 people and asked the question, How much cash are you carrying in your wallet? The mean of his data is $8.00 with a standard deviation of $2.50. Assume the data is normally distributed. Based on this information, answer the following questions. Part A Question Select the correct answer. What is the probability that a person who was surveyed has less than $5 in his or her wallet? Use the normal distribution table to help you in your calculations. 11.51% 21.19% 78.81% 88.49% Part B Question Select the correct answer. What is the probability that a person who was surveyed has between $9 and $10 in his or her wallet? Use the normal distribution table. 13.27% 31.04% 41.34% 86.73% Part C Now, Martin can reasonably guess that the standard deviation for the entire population of people at the mall during the time of the survey is $1.50. What is the 95% confidence interval about the sample mean? Interpret what this means in the context of the situation where 95 people were surveyed and the sample mean is $8. Use the information in this resource to help construct the confidence interval. Font Sizes Characters used: 0 / 15000 Part D Would the interval found in part C increase, decrease, or remain the same if the confidence level desired were 99%? State your reasoning. Font Sizes Characters used: 0 / 15000 Part E Would the interval found in part C increase, decrease, or remain the same if fewer than 95 people were surveyed? Justify your answer. Font Sizes Characters used: 0 / 15000

Answer 1

Answer:

Step-by-step explanation:

Sample size = 95

X=cash carried by the persons

x bar = 8.00

s = sample std dev = 2.50

Std error = $\frac{s}{\sqrt{n} } =\frac{2.5}{\sqrt{95} } \\=0.2565$

Hence Z score would be

$\frac{x-8}{0.2565}$

$a) P(X<5) = P(Z<\frac{x-8}{0.2565})\\=P(Z<-11.696)$

-0.00

b) $P(9<x<10) = P(3.89<Z<7.78)\\=0.00$

c) 95% conf interval margin of error = ±$1.96*0.2565$

=±0.54782

Confi interval = (8-0.5027, 8+0.5027)

= (7.4923, 8.5027)

C)If conf level increases, then width of interval would increase since critical value would increase.

If sample size increases std error would decrease and hence margin of error.

So interval would decrease.