can you check my work? What conditions (and theorem) must be met in order for a sampling distribution of means to be approximated by a normal distribution? answer: the central limit theorem population distribution is normal population distribution is symmetric, unimodal, without outliers, and the sample size is 15 or less population distribution is moderately skewed, unimodal, without outliers, and the sample size is between 16 and 40 sample size is greater than 40, without outliers
@amistre64
@Michele_Laino
I'm sorry I don't remember the central limit theorem
can you check a different problem?
ok!
An SRS of 400 American adults is asked, "What do you think is the most serious problem facing our schools?" Suppose that in fact 30% of all adults would answer drugs if asked this question. The proportion p-hat of the sample who answer drugs will vary in repeated sampling. In fact, we can assign probabilities to values of p-hat using the normal density curve with mean 0.3 and standard deviation 0.023. Use this density curve to find the probabilities of the following events: a) At least half of the sample believes that drugs are the schools' most serious problem. b) Less than 25% of the sample believes that drugs are the most serious problem. c) The sample proportion is between 0.25 and 0.35. a)normalcdf(0.5,1,0.3,0.023)=1.74712531^-18 P(X ≥ .5) = P((X - 0.3)/0.023 ≥ (.5 - .3)/.023) = P(Z ≥ 8.69) = P(Z ≤ - 8.69) = 0 b) normalcdf(0,.25,.3,.023)=.0148557801 P(X ≤ .25) = P((X - .3)/.023 ≤ (.25 - .3)/.023) = P(Z ≤ - 2.17) = .015003 c) P(.25 ≤ X ≤ .35) = P((.25 - .3)/.023 ≤ (X - .3)/.023 ≤ (.35 - .3)/.023) = P( -2.17 ≤ Z ≤ 2.17) = P(Z ≤ 2.17) - P(Z ≤ - 2.17) = .984997 - .015003 = .969994
Please wait I look at my tables...
okay
answer c. is right
what about a and b
answer b. 50-48.5= 1.5%
answer a. is right, namely the requested probability is zero
since we are very far from the mean
okay thank you!
Join our real-time social learning platform and learn together with your friends!