Find $P(Z>1.8)$ and $P(X>4)$ when $Z$ is normal RV and $X$ is a binomial RV.

Question

Question 1
If $Z\sim N(0,1)$, Find
a) $P(Z>1.8)$
b) $P(Z >-2.46)$
c) $P(Z>2.589)$
d) $P(Z<-0.725)$
e) $P(Z<-1.63)$
f) $P(Z>-0.65)$

My Solutions:
$P(Z>1.8) = 0.9461$
$P(Z >-2.46)= 1-0.99305 = 0.00695$
$P(Z>2.589) = 1-0.9952= 0.0048$
$P(Z<-0.725)= 1-0.7657=0.2343$
$P(Z<-1.63) = 1-0.9484 = 0.0516$
$P(Z>0.65) = 1-0.7422 = 0.2578$

Can you please check these over for me?

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Question 2
The random variable $X\sim\text{Bin}(6,0.45)$. Find
$P(X>4)$
$P(X=6)$
$P(X≤3)$

Can someone please help me with question 2? I tried researching and everything and still can't find a solution to answering this question.

Thank you.

Answer 1

Under your solutions for the first part, a) and b) are wrong.

Instead of research have you tried calculating anything?

If $X\sim\text{Binomil}(n, p)$, then the probability that $X = k$, for some integer $k$, $0\leq k\leq n$, is $$P(X = k) = \binom{n}{k} p^{k}(1-p)^{n-k}.$$

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The possible values for your $X$ are $\{0,1,2,3,4,5,6\}$.

Notice that, for $P(X>4)$, $$\{X>4\}\iff\{X=5\cup X = 6\}$$ and that the events $X=5$ and $X = 6$ are disjoint. Hence $$P(X > 4) = P(X=5\cup X = 6) = P(X = 5)+P(X=6).$$

Further, for $P(X\leq 3)$, $$\{X\leq 3\}\iff\{X = 0\cup X = 1\cup X=2\cup X=3\}.$$

You could also notice that $P(X\leq 3)+P(X = 4)+P(X>4) = 1$.

Answer 2

1. I suppose you are answering these questions using printed normal tables.

Notice that questions of the type $P(Z \le c)$ with $ c > 0$ can be read directly from most printed tables. For a continuous distribution such as the standard normal $P(Z < c) = (Z \le c).$ Questions involving negative $c$ require the use of symmetry, unless your book has tables that include negative values of $c$.

It is also possible to answer such questions using software and some models of statistical calculators. Here are answers to a few of the parts of this problem from R statistical software, including the first two parts, which @IanMiller has said you answered incorrectly, and one that requires averaging two values from a table. Notice that these answers are given to more places of accuracy than you will get from a printed table. In R software pbinom denotes the CDF of a normal distribution.

 1 - pnorm(1.8)   #a
 ## 0.03593032
 1 - pnorm(-2.46) #b
 ## 0.9930531
 pnorm(-.725)     #d average of the following two answers
 ## 0.234226
 pnorm(-.72)
 ## 0.2357625
 pnorm(-.73)
 ## 0.2326951

2. I suppose you are supposed to use the binomial formula for these. The Answer of @probablyme discusses use of the formula and gives you a good start.(+1)

I will give answers from R. You can use them to check your methods. Here, pbinom denotes the CDF and pbinom denotes the PDF, but you have to include parameters $n$ and $p$ as shown. How many decimal places are $you$ expected to show?

 1 - pbinom(4, 6, .45)
 ## 0.06919805
 sum(dbinom(5:6, 6, .45))  # '5:6' includes both 5 and 6
 ## 0.06919805
 dbinom(5, 6, .45) + dbinom(6, 6, .45)
 ## 0.06919805

 dbinom(6, 6, .45)
 ## 0.008303766
 .45^6
 ## 0.008303766

 pbinom(3, 6, .45)
 ## 0.7447361
 sum(dbinom(0:3, 6, .45))
 ## 0.7447361

Note: You should probably investigate using software in this course, if it is not already part of the course. R software that works on a wide variety of computers is available free of charge from r-project.org. There is a lot to learn about R; just focus on the parts you need. If available, many other kinds of statistical software would do as well.