Question

Thirty percent of all people believe lying is okay in certain situations. A group of 50 people were sampled and asked the same question.

a) describe the sampling distribution

b) Find the probability that less than 20 of the sample are okay with lying.

c) What proportion represents the top 30% chance of occurring?

Answer 1

Given,

Thirty percent of all people believe lying is okay in certain situations.

i.e

Proportion of people believe lying is ok : p = 0.30

A group of 50 people were sampled and asked the same question

Sample size : n= 50

a)

Samping distribution of Sample proportion of people believe lying is okay in certain situations: follows normal distribution with mean = p = 0.30 and standard deviation :

b)

probability that less than 20 of the sample are okay with lying.

Number of people believe lying is okay in the sample = 20

Sample proportion =20/50 =0.4

probability that less than 20 of the sample are okay with lying = P(<0.4)

Z-score for 0.4 = (0.4-0.3)/0.0648=0.1/0.0648 = 1.54

From standard normal tables, P(Z<1.54) = 0.9382

P(<0.4) = P(Z<1.54) = 0.9382

probability that less than 20 of the sample are okay with lying = 0.9382

c) What proportion represents the top 30% chance of occurring?

Let p30 be the proportion represents the top 30% chance of occurring

i.e

P(>p30) = 0.30

P(>p30) = 1 - P(<p30) = 0.30

P(<p30) =1-0.30=0.70

Z30 be the Z-score for p30

i.e Z30 = (p30-0.30)/0.0648

p30 = 0.30+0.0648Z30

From standard normal tables,'

P(Z < 0.52) = 0.6985

P(Z<0.53) = 0.7019

ThereforeZo= 0.525 is good approximation for P(Z<Z0) = 0.70

p30 = 0.30+0.0648Z30 =0.30 + 0.0648 x 0.525=0.30+0.03402=0.33402

p30 = 0.33402

proportion that represents the top 30% chance of occurring= 0.33402