A sample of 6 Tiger sharks has a mean length of 13.72 feet with a standard deviation of 2.66 feet. Assume the lengths are normally distributed. Find a 80 % confidence interval for the standard deviation in the lengths of Tiger Sharks. x = 1.610 x² = 9.236 Sos (Round to 3 decimal places) (Round to 3 decimal places) (Round to 3 decimal places)

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A: Gievn that:Sample size: Confidence level:

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A: Mean(µ) = 49Standard deviations (σ) = 16

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A: 18.08,15.79,16.45,20.42,20.88,15.65,17.09,15.38,15.64,19.56,16.84,16.95,19.31,18.69,21.66,17.14

Q: For a standard normal distribution, find: P(Z > 2.47)

A: P(z>2.47)

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A:

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A:

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A: Sample size Mean Standard deviation NOTE: Hi there! Thanks for posting the question. Since you have…

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A: There are total 4 groups, each containing 12 respondents.

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A: Sample proportion p^ =0.26Margin of error=0.08

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A: From data Minimum value= 15.9Maximum value= 38.9

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A: Data: 1.6,1.5,2.6,2.3,2.3,2.3,2.6,3,1.5,2.3Note:- According to the Bartleby guidelines experts solve…

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A: The given data is as follows:

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A: NumberProbability00.3610.2620.1930.084 or more0.11

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A: Population mean=992Population standard deviation=141Sample size n=37

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A: It is given that favourable cases, x = 19Sample size, n = 19 + 87 = 106

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A: we have sort the data from smallest to largest: 1818212223242526272930313232323334353535363739

Q: 3. The transition matrix for a three-state Markov chain is 0 ||1 0 r llo q p+r Р P = q Where p, q>…

A:

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A: Note: As per our guidelines we are supposed to answer only 3 sub-parts of any questions so I am…

Q: percentage of scores lie within 2.34 standard deviations on both sides of the mean?

A: Standard normal distribution

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Q: 27 Solar energy cells. According to Energy Trend (January 2019), 38% of the solar energy cells…

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Q: Find the probability that the proportion of freshmen in the sample of 130 who plan to major in a…

A: It is given that p = 27% = 0.27n = 130

Q: Starting salaries of 64 college graduates who have taken a statistics course have a mean of $43,500…

A: Sample size n=64Sample mean=43500Standard deviation=6800

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A: Confidence interval ( 12.08 &lt;  &lt; 12.72) LCL=12.08UCL=12.72

Q: Determine the sample size required to estimate a population mean to within 11 units given that the…

A: E=11,σ=50,CI=95%=0.95α=1−CI=1−0.95=0.05

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A: At 95% critical value of z is 1.96. . From z table

Q: 5. [Section 4.9] The number of traffic accidents in Berkeley in 10 randomly chosen non-rainy days in…

A: Data 4,0, 6, 5, 2, 1, 2, 0, 4, 3

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A: Sample size (n) = 32Mean x¯ = 20,720 BTUsStandard deviation (s) = 270 BTUsSignificance level(α) = 1…