Question

Insurance company records indicate that​ 14% of its policyholders file claims involving theft or robbery of personal property from their homes. Suppose a random sample of 500 policyholders is selected. What is the standard deviation of the sampling distribution of the sample proportion of policyholders filing claims involving theft or robbery from their​ homes?

Answer 1

Answer: std dev = 0.01552

Step-by-step explanation:

we will solve this by taking a step by step analysis of the problem.

i hope at the end of this section, you will feel confident  to try out other questions.

given that the size of the sample (n) = 500

The Insurance company records indicate that 14% of its policyholders file claims involving theft or robbery of personal property from their homes.

hence sample proportion of policyholders filing claims involving theft or robbery from their homes (p) = 14% =0.14

let Z be the number of policyholders files involving theft or robbery of personal property from their homes.

so Z~Bin(500,0.14) and p=Z/500

we know from basic mathematics that standard deviation = √(variance)

so variance is  given thus;

V[Z] = 500*0.14*(1-0.14)

or V[p] = V[Z/500] = 500*0.14*0.86/500² = 0.14*0.86/500

so standard deviation is;

sqrt[V[p]]=sqrt(0.14*0.86/500) = sqrt(0.0002408) = 0.01552

∴ the standard deviation = 0.01552

cheers i hope this helps!!!!

Answer 2

Answer:

Standard deviation = 7.76

Step-by-step explanation:

The formula for determining standard deviation when dealing with population proportion is expressed as

Standard deviation = √npq

Where

n represents the number of samples from the population.

p represents the probability of success.

q represents the probability of failure.

From the information given,

n = 500 policyholders

p = 14% = 14/100 = 0.14

q = 1 - 0.14 = 0.86

Standard deviation = √500 × 0.14 × 0.86 = 7.76