Question

Be sure to carefully follow the instructions about entering your answer. In most open-answer problems, only a number is required (i.e., no units, no percentage sign, etc...). Pay close attention to the number of decimal places requested. You should use a TI-84 PLUS calculator for all procedures where appropriate. Question 5 (5 points) A survey of high school students revealed that the numbers of soft drinks consumed per month was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average number of soft drinks consumed per month for the sample was between 29.5 and 30 soft drinks? Write only a number as your answer. Round to 4 decimal places (for example 0.0048). Do not write as a percentage. Your Answer: Answer Page 5 of 20 Next Page Previous Page

Answer 1

Solutions:-
5. Given that mean = 25, and standard deviation = 15, n = 36

P(29.5 < X < 30)
= P((29.5-25)/(15/sqrt(36)) < (X-mean)/(s/sqrt(n)) < (30-25)/(15/sqrt(36))
= P(1.8 < Z < 2)
= 0.0131

4. Given that 287,311,262,392,313,267,303,316,286,310,263,291
n =12, X = 300.08 , σ = 51
95% Confidence interval for Z = 1.96

95% Confidence interval for the population mean = X +/- Z*σ/sqrt(n)
= 300.08 +/- 1.96*51/sqrt(12)
= (271.2 , 328.9)

3. Given that mean = 11.5 σ = 2.7, n = 36, X = 11.26

P(X < 11.26) = P(Z < (11.26-11.5)/(2.7/sqrt(36)))
= P(Z < -0.5333)
= 0.2981