In performing a chi-square goodness-of-fit test for a normal distribution, a researcher wants to make sure that all of the expected cell frequencies are at least five. the sample is divided into 7 intervals. the second through the sixth intervals all have expected cell frequencies of at least five. the first and the last intervals have expected cell frequencies of 1.5 each. after adjusting the number of intervals, the degrees of freedom for the chi-square statistic is

2, 3,5, or 7

left 3, up 4

step-by-step explanation:

just count the values of x & y of the vertex of the graph and see how many units in the x and y direction will get you from the f(x) graph to the f(x) graph

answer: do we write it

step-by-step explanation:

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %