38% of working mothers do not have enough money to cover their health insurance deductibles. You randomly select six working mothers and ask them whether they have enough money to cover their health insurance deductibles. The random variable represents the number of working mothers who do not have enough money to cover their health insurance deductibles. Complete parts (a) through (c) below.
(a) Construct a binomial distribution using n equals 6n=6 and p equals 0.38p=0.38.
x. p(x)
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2. ?
3. ?
4. ?
5. ?
6. ?
38% of working mothers do not have enough money to cover their health insurance deductibles. You...
equen 36% of working mothers do not have enough money to cover their health insurance deductibles. You randomly select six working mothers and ask them whether they have enough money to cover their health insurance deductibles. The random variable represents the number of working mothers who do not have enough money to their health insurance deductibles. Complete parts (a) through (c) below (a) Construct a binomial distribution using n 6 and p 0.36 P(x 0 2 For H 4 5...
36% of working mothers do not have enough money to cover their health insurance deductibles. You randomly select six working mothers and ask them whether they have enough money to cover their health insurance deductibles. The random variable represents the number of working mothers who do not have enough money to cover their health insurance deductibles. Complete parts (a) through (c) below. (a) Construct a binomial distribution using n=6 and p=0.36. х P(x) 0 1 2 3 4 5 6...
Question Help bo Triya va 32% of working mothers do not have enough money to cover their health insurance deductibles. You randomly select six working mothers and ask them whether they have enough money to cover their health insurance deductibles. The random variable represents the number of working mothers who do not have enough money to cover their health insurance deductibles. Complete parts (a) through (c) below. (a) Construct a binomial distribution using n=6 and p=0.32. ire Ag P(x) om...
A survey of 1 comma 567 randomly selected adults showed that 570 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 38% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (d. Sample proportion: 0.363752 Test statistic, z: negative 1.3251 Critical z: plus or minus1.9600 P-Value:...
I'm working on probability HW and working on word problems. Do you have any tips as to how to differentiate between Poisson distributions from Binomials, Geometric and Negative Binomials when doing word problems...i provided an examples for everything but Possion, please explain Poisson distribution using the same example structures Binomial is simplest one in this we have number of trials given 'n' and probability of success 'p' and they ask for probability of getting success 'r' times. For example: When...
a survey of 1017 adults, a polling agency asked, "When you retire, do you think you will have enough money to live comfortably or not the 1017 surveyed, 5 ated that they were worried about having enough money to live comfortably in retirement Construct a 95% confidence interval for the proportion of adults who ar orried about having enough money to live comfortably in retirement lick here to view the standard normal distribution table (page 1) lick here to view...
The second photo is an example of the equation I am expected
to use. It says use technology but I have no idea how to do
that.
Coumes Assignments&Homework P Do Homeswork Mathew McC Determine whether youcan u x https:/www.mathxl.com/Student/PlayerHomework.aspx?homeworkd-5305724008questiond-5&ushedasecid-55113768centerwinyes Elementary Statistics: Picturing the World 7/e-Internet-Summer 1 2019-Mays 7/4/19730 AM Mathew McCarley & Homework: CH 5.5 Normal Approximations to Binomial Distribut Save Score: 0 of 1 pt 7 of 76 complete) HW Score: 69 27 % , 5.54 of 8...
Suppose that insurance companies did a survey. They randomly
surveyed 410 drivers and found that 300 claimed they always buckle
up. We are interested in the population proportion of drivers who
claim they always buckle up.
NOTE: If you are using a Student's t-distribution, you may
assume that the underlying population is normally distributed. (In
general, you must first prove that assumption, though.)
Part (a)
(i) Enter an exact number as an integer, fraction, or decimal.
x =
(ii) Enter...
Have to show work for every problem
4. A company uses three plants to produce a new computer chip. Plant A produces 30% of the chips. Plant B produces 45% of the chips. The rest of the chips are produced by plant C. Each plant has its own defectiv rate. These are: plant A produces 3% defective chips, plant B produces 1% defective chips, plant C produces 5% defective chips. Hint: draw a tree diagram. (a) Construct a tree diagram...
1. Many companies use a incoming shipments of parts, raw materials, and so on. In the electronics industry, component parts are commonly shipped from suppliers in large lots. Inspection of a sample of n components can be viewed as the n trials of a binomial experimem. The outcome for each component tested (trialD will be that the component is classified as good or defective defective components in the lot do not exceed 1 %. Suppose a random sample of fiver...