also, instead of doing the figure he did to find how many standard deviations from the mean k is, you can use z=(x-μ)/σ, where x is 52 in the first case, μ is 60 and σ is 4. If you do that, you'll get -2, and if you do that again with x=68, you'll get 2. So we need +/-2 std deviations, and k =2. Maybe this will be easier for someone than drawing a figure.
As long as there are 20 values. When there are an even number of values, there is no single middle value. For example, if you have 3 values, say 1, 4, 6, then 4 (the second value) is the middle value. But if you have 4 values (an even number of values), say 4, 5, 8, 9, there is no single middle value. Instead we can think of it as having the middle value between the second and third value. Similarly, if there are 20 values, the middle value can be thought of as being between the 10th and 11th values.
S Behrens The Bernoulli distribution is a special case of the binomial distribution where only a single trial is conducted (that is, Binomial with n=1).
Then what if it has this??? _________ {A n B} it has the long line on the top which is a compliment but how is this solved??? A= 1,2,3 B= 4, 5, 6 U= 7, 8, 9,
