Question

1. Binomial Experiment application:
   1. In order to be considered for a tier 1 point guard in women’s basketball, you need to be 5’8”. Let’s consider a “success” as finding a woman who is 5’8” or taller. The probability of success is approximately 5%. If we consider a sample of 1000 women, calculate the mean and the standard deviation of the binomial random variable.
   2. Calculate the range of “normal” observations (not unusual) for the number of women that we would expect to be 5’8” or taller in a sample of 1000 women and use it in a sentence.
   3. Use a binomial calculator to calculate the probability of selecting*:
      1. Less than 40 women who are 5’8” or taller
      2. Exactly 60 women who are 5’8” or taller
      3. Between 50 and 80, inclusive, women who are 5’8” or taller
      4. More than 70 women who are 5’8” or taller
      5. At least 60 women who are 5’8” or taller

*Don’t just give answers here, use complete sentences. Graphs would be a nice addition here (StatCrunch screen grabs or Excel graphs).

4. Can we use the normal distribution as an approximation for the binomial in this case? Why or why not? If yes, what is the probability that we would choose less than 40 women who are 5’8” or taller using this approximation? How does this value compare with the value calculated in part c(i) above?

Answer 1

Let the probability of success (as finding a woman who is 5’8” or taller) be p. Let the random variable representing this binomial distribution be X (no. of successes in n no. of attempts)

p = 0.05

A.

n = 1000

Mean of a normal distribution, = n*p

= 1000*0.05

= 50

Std. Deviation =

= 6.892

B.

Quartile 1, Q1:

P(Xx) = 0.25

x = 44.77

~ 45

Quartile 3, Q3:

P(Xx) = 0.75

x = 54.064

~ 55

I have rounded up the numbers because a binomial random variable can only be an integer.

Interquartile range, IQR = Q3-Q1

= 55-45

= 10

Lower Outlier Boundary = Q1-1.5*IQR

= 45-1.5*10

= 30

Upper Outlier Boundary = Q3+1.5*IQR

= 55+1.5*10

= 70

So, the normal observation range is: [30,70]

Thus, a sample won't be unusual if there are 30 to 70 women that are 5’8” or taller in a sample of 1000 women. Anything outside this interval would not be normal.

Note: I have used soft outlier boundaries (used the factor of 1.5*IQR, generally this is used to report outliers). For hard outlier boundaries, we should use a factor of 3*IQR.

C.

We know that,

P(X=x) = (nCx)*p^x*(1-p)^n-x

Using the above formula:

i.

P(X<40) = 0.0598

ii.

P(X = 60) = 0.01967

iii.

P(50X80) = P(X80) - P(X<50)

= 0.999965 - 0.47974

= 0.520225

iv.

P(X60) = 0.086732

D.

We know that a binomial distribution can be approximated with a normal distribution if,

n*p and n*(1-p) > 5

n*p = 50

n*(1-p) = 950

Since both of these conditions are met, we can approximate x by a normal distribution with

mean, = n*p

= 50

and

Std. deviation, =
= 6.892

As we know that binomial is a discrete distribution and normal is a continuous one so we have to do some continuity correction before calculating probability using a normal approximation.

Thus,

if, P(X<n) = P(X<n-0.5)

So,

P(X<40) = P(X<39.5)

We know that an std. normal variable Z is:

Z = (X-)/

So,

P(X<40) = P(X<39.5) = P(Z<(39.5-50)/6.892)

= P(<-1.5235)

= 0.0638

Answer in part c(i):

P(X<40) = 0.0598

Difference = 0.0638-0.0598

= 0.004

Percentage = 0.004*100/0.0598

= 6.6%

Thus difference is not very big, hence approximation seems valid.

Please upvote if you have liked my answer, would be of great help. Thank you.