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N represents the total number of trials. The number I’ve chosen to work with is 7 X the projected number of successes. Because the number of successes can never be larger than the number of tries in any condition, x must be less than N, which in our instance is 7. As a result, we can use 5 as our x.
P denotes the trial’s success probability. Considering the question, we must keep P = 0.6.
As a result, for N=7, x=5, and p=0.6, the probability is 0.261.
Normal Distribution (Part 2)
Question: select your own values of z1 and z2, then use the Appendix Table for the Standard Normal
Our value of choice for z1 and z2 are: z1 = -2.47
z2 =-1.98
Distribution to find:
1) P (z < z1)
Replacing z1 in the equation with -2.47 which is the value of choice for z2
P (z < -2.47)
Finding the probability from the table we get the solution to be
P (z < -2.47) = 0.00676
= 0.00676
2) P (z > z1)
Replacing z1 in the equation with -2.47 which is the value of choice for z1
P (z > -2.47)
Finding the probability from the table we get the solution to be
P (z > -2.47) = 1 - P (z > -2.47)
= 1- 0.00676
= 0.9932
3) P (z < z2)
Replacing z2 in the equation with -1.98 which is the value of choice for z2
P (z < -1.98)
Finding the probability from the table we get the solution to be
P (z < -1.98) = 0.02385
= 0.02385
4) P (z > z2)
Replacing z2 in this equation with -1.98 which is the value of choice for z2
P (z > -1.98)
Finding the probability from the table we get the solution to be
P (z > -1.98) = 1 - P (z < -1.98)
1 - 0.02385 = 0.9762
= 0.9762
5) P (z1 < z < z2)
Replacing for both z1 and z2 in the equation with z1 = -2.47 and z2 = 1.98 representing the values we choose to work with. The equation P (z1 < z < z2) therefore can be written as:
P (-2.47 < z < -1.98)
Which can be broken and written in a simpler way as
P (z < -1.98) - P (z < -2.47)
Recall that from the above solutions P (z < -1.98) = 0.02385 and P (z < -2.47) = 0.00676.
Therefore, the solution is:
P (z < -1.98) - P (z < -2.47)
= 0.02385 - 0.00676
= 0.0171
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