A certain type of light bulb has an average life of 1100 hours, with a standard deviation of 50 hours. The length of life of the bulb can be closely approximated by a normal curve. An amusement park buys and installs 80.000 such bulbs. Find the total number that can be expected to last at least 1100 hours The number of light bulbs that can be expected to last at least 1100 hours s

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A:

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A: Solution: Jim is going to build a playlist that contains a 14 songs.

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A: Given P=0.70 q=0.30

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A: Given Information: Standard deviations: σ1=9, σ2=5 Sample sizes n1=10, n2=14 Sample means: x¯1=4.5,…

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A: given data math : μ1 = 63     σ1  = 7biology : μ2 = 23      σ2 = 3.9

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A: given data claim : p ≠0.000340n = 420143x = 130p^ = xn = 130420143 = 0.000309

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A: A recent poll indicates that 48% of all registered voters in America plan to vote for a specific…

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A:

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A: given data, 151- yellow peas415- green peas 566- toatal peas α=0.05claim :p=0.25we have to complete…

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A: Givensample size(n)=196Mean(x)=47.78standard deviation(s)=3.85

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A: The following information has been provided: Placebo  Melatonin Gummy  Prescription 5 2 7 1 6…

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A: Given that  Total cards = 52 Black cards = 26 Red cards= 26

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A: Given N=244 X=55 Alpha=0.10 P=0.2

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A: From the provided information, Sample size (n) = 16 Level of significance (α) = 0.10 Population…

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A: According to given question,  We have :-  Sample size n= 40 Sample standard deviation s  =  13.1…

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A: Machine 1:Sample size, n1=15Sample mean diameter, x1=8.69The sample variances, s12=0.35Machine…

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A: Given: Percentage of making free throws is 87% p=0.87 therefore q=1-p=1-0.87=0.13

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A: given data, sample proportion p^=0.59 2E=0.3E=0.32=1.5CI=0.99α=1-0.99=0.01we have to find out the…

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Q: Test the claim at the 5% significance level that a suitable model for these data is a Poisson…

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A: From the given information, The null hypothesis is,Ho:μ1=μ2 The alternative hypothesis is, H1:μ1≠μ2…

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A: x f fx fx2 210 3 630 132300 370 4 1480 547600 530 3 1590 842700 740 1 740 547600 1140 1…

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A: given data  The test statistic of z=−3.02 is obtained when testing the claim that p=2/3.

Q: 5. Let X₁, X2, X3, X4, X5 be a random sample of size 5 from the pdf 2x 82 ₁0 0. Suppose we want to…

A: It is given that X1, X2, …X5 be a random sample that follows the pdf  fx;θ=2xθ2, 0&lt;x&lt;θ The…

Q: Example 14 In a sample survey of the safety explosives' used in certain mining operations,…

A:

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A: R programming has used to calculate the probability area.

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A: given data, n=100p(blue)=p=0.24claim:p=0.21α=0.01

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A: “Since you have posted multiple questions, we will provide the solution only to the first question…

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A: The degrees of freedom for the sum of squares explained by the regression is the number of…

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A: Given: Two new drugs were given to patients with heart disease . The first drug lowered theblood…

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A: given data n1 = 9n2 = 9α = 0.05we have to find critical value for two tailed test.

Q: A random sample of 838 births included 429 boys. Use a 0.01 significance level to test the claim…

A: given data claim : p =0.513n = 838x = 429α = 0.01p^ = xn = 429838 = 0.5119

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A: Given information: Claim: 3% of users develop nausea. Null Hypothesis: H0: p=0.03 Alternative…

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A: Given,no.of defective batteries=1no.of non defective batteries=11Total no.of batteries=12

Q: If a set of sample measurements has a mean of 100, a normal distribution, a standard deviation of 2,…

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Q: Assume that a procedure yields a binomial distribution with n = 763 trials and the probability of…

A: From given data we have : sample size = n = 763 probability of success = p = ( 4 / 15 ) =0.2667