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The lengths (in minutes) of a random selection of popular children’s animated films are

listed below. Estimate the true mean length of all children’s animated films with 95%

confidence.

90 84 83 91 75 88 78 96 78 79 77


2. Changes in airport procedures require considerable planning. Arrival rates of aircraft are important factors that must be taken into account. Suppose small aircraft arrive at a certain airport, according to a Poisson process, at the rate of 5 per hour. Thus, the Poisson parameter for arrivals over a period of hours is μ = 5t.

(a) What is the probability that exactly 4 small aircraft arrive during a 1-hour period?

(b) What is the probability that at least 4 arrive during a 1-hour period?

(c) If we define a working day as 12 hours, what is the probability that at least 75 small aircraft arrive during a working day?


1.The average number of field mice per acre in a 5-acre wheat field is estimated to be 10. Find the probability that fewer than 6 field mice are found

(a) on a given acre;

(b) on 2 of the next 3 acres inspected.


Directions: From the given situation, (a) state the null and alternative hypotheses, (b) compute the test statistic, (c) determine the critical value and sketch the rejection region and non-rejection region in the normal curve.



1. The cahier of a fast-food restaurant claims that the average amount spent by customers for dinner is Php 120.00. A sample of 50 customers over a month was randomly selected and it was found out that the average amount spent for dinner was Php 122.50. Construct the critical regions using a 0.05 level of significance to conclude that the average amount spent by customers is more than Php 120.00. Assume that the population standard deviation is Php6.50.



Activity 1: Draw Me


Directions: Given the following information, construct the rejection region. Show the solution in a step-by-step procedure.


1. H0 : = 84


Ha : 84


m= 87, s= 10, n = 35, alpha= 0.05



2. H0 : = 45


Ha : < 45


m= 40, s = 12, n = 32, alpha= 0.01


The average zone of inhibition (in mm) for mouthwash L as tested by the medical technology students has been known to be 9mm. A random sample of 10 mouthwash L was tested and the test yielded an average zone of inhibition of 7.5mm with a variance of 25 mm. Is there enough reason to believe that the anti-bacterial property of the mouthwash has decreased? Test the hypothesis that the average zone of inhibition of the mouthwash is no less than 9mm using 0.05 level of significance.


A. State the hypotheses.


B. Determine the test statistic to use.


C. Determine the level of significance, critical value, and the decision rule.


D. Compute the value of the test statistic.


E. Make a decision.


F. Draw a conclusion.

If the average time spent studying is 10 hours and the standard deviation is 4 hours, what it the probability that a student will spend more than 13 hours studying?


An experimental study wad conducted by a researcher to determine if a new time slot has an effect on the performance of pupils in mathematics . Fifteen randomly selected learners participated in the study . Toward the end of the investigation , a standardized asssesment was conducted. The sample mean x= 75 and s=5 . In standardization of the test, the mean was 65 and standard deviation was 8. Based on the evidence at hand, is the new time slot effective ? Use a= 0.05.

A study believes that 70% of adults in the Philippines own a cellphone.

A cellphone manufacturer believes that the actual number is much

less than 70%. 100 Filipino adults were surveyed, of which 74 have

cellphones. Using a 5% level of significance, is the cellphone

manufacturer’s claim valid or not?


The letters in the word TRIANGLE are to be arranged. How many ways can these letters be arranged in a line such that letters E and G appear in adjacent positions?

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