Continuity Corrections CS1 Exam paper

[SYC]

Hi everyone,

Sorry if this is quite easy. I've started doing a paper September 2019 and was stuck on question 1.

The question:

A Survey showed that 40% of investors invest in at least two companies in order to diversify their risk.
Calculate an approximate probability that more than 100 investors have invested in at least two companies in a random sample of 300 investors.

I assumed it was a binomial distribution so the expected number and variance would align to the binomial ones but when it came to using the continuity corrections the mark scheme states:

1 - (phi)((100.5-120)/sqrt(72))
= 1 - (phi)(-2.298)
= (phi)(2.298)
= 0.989

My question is how did they get from 1 - (phi)(-2.298) to (phi)(2.298) ? where did the 1 go and how did -2.298 becomes positive (my assumption is the flip graph? as from the binomial distribution the probability is calculated using Z grid and starts from the mean, I'm confused why the P(X>(or equal)100.5) would be subtracted from 1 instead of adding 0.5 from the other half of the graph?

Thank you in advance!

 

[mugono]

Hi everyone,

Sorry if this is quite easy. I've started doing a paper September 2019 and was stuck on question 1.

The question:

A Survey showed that 40% of investors invest in at least two companies in order to diversify their risk.
Calculate an approximate probability that more than 100 investors have invested in at least two companies in a random sample of 300 investors.

I assumed it was a binomial distribution so the expected number and variance would align to the binomial ones but when it came to using the continuity corrections the mark scheme states:

1 - (phi)((100.5-120)/sqrt(72))
= 1 - (phi)(-2.298)
= (phi)(2.298)
= 0.989

My question is how did they get from 1 - (phi)(-2.298) to (phi)(2.298) ? where did the 1 go and how did -2.298 becomes positive (my assumption is the flip graph? as from the binomial distribution the probability is calculated using Z grid and starts from the mean, I'm confused why the P(X>(or equal)100.5) would be subtracted from 1 instead of adding 0.5 from the other half of the graph?

Thank you in advance!

Hi,

The question is asking you to calculate the approximate probability. The 'exact' probability would be determined using the binomial distribution. This would've been the approach to use had you not been asked to calculate the approximate probability.

Let X~Bin(n,p). Then the approximate distribution (say) Y ~ N(np, npq)
A continuity correction is required when moving from a discrete to a continuous distribution.
The P(X > 100) ~ P(Y > 100.5).

Standardising the distribution of Y to the standard normal with mean 0, and variance 1 gives

P(Z > [100.5 - 120] / sqrt(72)) = P(Z> -2.298)

Graphically, this probability represents the area under the curve to the right of -2.298 of a N(0,1) distribution.
Now, the normal distribution is a symmetric distribution. This property is desrable and means that P(Z>-2.298) = P(Z<2.298) = phi(2.298) [the area under the curve to the left of 2.298.]

Hope that helps.

 

[SYC]

Hi,

The question is asking you to calculate the approximate probability. The 'exact' probability would be determined using the binomial distribution. This would've been the approach to use had you not been asked to calculate the approximate probability.

Let X~Bin(n,p). Then the approximate distribution (say) Y ~ N(np, npq)
A continuity correction is required when moving from a discrete to a continuous distribution.
The P(X > 100) ~ P(Y > 100.5).

Standardising the distribution of Y to the standard normal with mean 0, and variance 1 gives

P(Z > [100.5 - 120] / sqrt(72)) = P(Z> -2.298)

Graphically, this probability represents the area under the curve to the right of -2.298 of a N(0,1) distribution.
Now, the normal distribution is a symmetric distribution. This property is desrable and means that P(Z>-2.298) = P(Z<2.298) = phi(2.298) [the area under the curve to the left of 2.298.]

Hope that helps.

Thank you so much Mugono! You've really cleared things up for me!