jacobdesalvo8155
17.12.2021 •
Mathematics
If I tell my kid "Go to bed" the amount of time it takes for her to actually go to bed is normally distributed with a standard deviation of 8.6 minutes. I want an 83% confidence interval to estimate the average time that it takes my 2.5 year old daughter to go to bed after saying "Go to bed" the first time. I collected the sample for 15 nights and the mean time over 15 nights is 29.18 minutes. What is the z-score I will need for my confidence interval? Use three decimal places lower end of the interval (use three decimal places): upper end of the interval (use three decimal places): Now assume that the standard deviation is not known and the sample standard deviation is 8.2 minutes. I will again construct a 83% confidence interval. Lower end of the interval (use three decimal places): Upper end of the interval (use three decimal places):
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Ответ:
Its c / b
Step-by-step explanation:
2a/b * 4c / 8a
= (2a*4c) / (b * 8a) Now 8a / 2a = 4 so:-
= 4c / 4b
= c / b