The distribution under study is X~Binomial(10,.91)
a) 0.0078 (P(X=6))
b) 0.9912(P(X>6)=1-P(X<=6))
c) 0.001(P(X<6)=P(X<=5))
d) 0.999 (P(X>=6)=1-P(X<=5))
A hotel claims that 91% of its customers are very satisfied with its service. Complete parts...
MUST use the Binomail Distribution function BINOM.DIST() function. Provide your final answer rounded to 4 decimal places in the yellow highlight cell. a) What is the probability that 6 customers are very satisfied? b) What is the probability that more than 6 customers are very satisfied? c) What is the probability that less than 6 customers are very satisfied? d) What is the probability that at least 6 customers are very satisfied? A hotel claims that 91% of its customers...
A hotel claims that 85% of its customers are very satisfied with its service. Complete parts a through d below based on a random sample of seven customers. a. What is the probability that exactly six customers are very satisfied? (Round to four decimal places as needed.) b. What is the probability that more than six customers are very satisfied? (Round to four decimal places as needed.) c. What is the probability that less than five customers are very satisfied?...
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A hotel claims that 85% of its customers are very satisfied with its service. Complete parts a through d below based on a random sample of five customers. a. What is the probability that exactly four customers are very satisfied? 0.3915 (Round to four decimal places as needed.) b. What is the probability that more than four customers are very satisfied? 0.4437 (Round to four...
A tourism company claims that 50% of its customers are satisfied with the service and prices. a. If this claim is true, what is the probability that in a random sample of 600 customers less than 45% are satisfied? b. Suppose that in a random sample of 600 customers, 270 express satisfaction with the company. What does this tell you about the company’s claim?
epolled on this issue, A consumer advocate claims that 85 percent of cable television subscribers are not satisfied vwith their cable service. In n attempt to justify this claim, a randomly selected sample of cable subscribers will nthe sample are not satisfied formula to compute the probability that 6 or more subscribers h i end Do not round intormediat ealuations Rund Fnal anewor to in 2 decimal place Bound other final angere to d decimal pics 85 Binomial, n. Probability...
A consumer advocate claims that 85 percent of cable television subscribers are not satisfied with their cable service. In an attempt to justify this claim, a randomly selected sample of cable subscribers will be polled on this issue. (a) Suppose that the advocate's claim is true, and suppose that a random sample of 8 cable subscribers is selected. Assuming independence, use an appropriate formula to compute the probability that 7 or more subscribers in the sample are not satisfied with...
A consumer advocate claims that 70 percent of cable television subscribers are not satisfied with their cable service. In an attempt to justify this claim, a randomly selected sample of cable subscribers will be polled on this issue. (a) Suppose that the advocate's claim is true, and suppose that a random sample of 4 cable subscribers is selected. Assuming independence, use an appropriate formula to compute the probability that 3 or more subscribers in the sample are not satisfied with...
Just question E In the Ardmore Hotel, 20% of the customers pay by American Express credit card. Of the next 10 customers, what is the probability that: a) None pay by American Express. b) At least two pay with American Express. c) Fewer than three pay with American Express. d) What is the expected number who pay by American Express? e) Using your understanding of the characteristics of a binomial experiment, explain why this is an example of a binomial...
If 82 out of 174 customers are somewhat or very satisfied with your customer service, then what is p, the sample proportion? If you make a 95% confidence interval, what is the parameter you are estimating? Oo Op Op (Rho, the lower-case Greek r) If you want to compute a 95% confidence interval for the parameter, what are the assumptions that must be true? Othe original variable is normally distributed On2 30 On*p is large AND n*(1-p) is large (at...
An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by the local dealer. The customer relations department will survey a random sample of customers and compute a 96% confidence interval for the proportion who are not satisfied. Past studies suggest that this proportion will be about 0.29. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.01. (You will need...