Answer:
Step-by-step explanation:
Hello!
Be it a variable with a binomial distribution, if n≥30, n*p≥5 and n*q≥5, then you can approximate the distribution of the sampling proportion to normal:
^p≈N(p;(p*q)/n)
Where p is the probability of success and q is the probability of failure (q= 1-p)
Under this distribution you can calculate the probabilities using the standard normal approximation:
[tex]Z=\frac{p(hut)-p}{\sqrt{\frac{p*q}{n} } }[/tex]≈N(0;1)
Find P(0.06 < p-hat < 0.16)=_______
To find the probability inside the given interval you have to subtract the accumulated up to the lower bond to whats accumulated up to the upper bound of the interval:
P(^p<0.16)-P(^p<0.06)
Now you standardize both terms to obtain Z-values and look for the corresponding probabilities in the Z-table:
P(Z<(0.16-0.10)/(√(0.10*0.9)/200))-P(Z<(0.06-0.10)/(√(0.10*0.9)/200))
P(Z<2.82)-P(Z<-1.89)
Now what's left is to look for the values in the Z-table. All negative values are in the left entry and all positive values are in the right entry. The integer and first decimal can be found in the first column of the table, the second decimal can be found in the first row of the table when you cross both, you find the corresponding cumulative probability.
P(Z<2.82)-P(Z<-1.89)= 0.99760-0.02938= 0.96822
I hope this helps!
