Question

. A department store has determined that 36% of all their sales are credit sales. A random sample of 65 sales is selected.

(a) Explain the distribution of the sample proportion of the credit card sales.

(b) What is the probability that the sample proportion will be greater than 0.40?

(c) What is the probability that the sample proportion will be between 0.30 and 0.40?

(d) What is the probability that the sample proportion will be less than 0.3

Answer 1

1)Hence probability that sample proportion greater than 0.4 is 0.2937

2)Hence the probability that sample proportion is between in 0.30 and 0.40 is 0.6607

3)Hence the probability that getting the sample proportion less that 0.3 is 0.1898.

Explanation:

Given:

36 % of sales are credit sales.

And 65 sales are randomly selected.

To Find:

1) Explain the distribution of the sample proportion of the credit card sales.

2) What is the probability that the sample proportion will be greater than 0.40?

3) What is the probability that the sample proportion will be between 0.30 and 0.40?

4) What is the probability that the sample proportion will be less than 0.3

Solution:

For a sample randomly selected of size "n" from total population,sampling distribution of sampling proportion is normal when sample size is large.]

Sample distribution follows normal with mean =p

here p=0.36 ,n=65

And

`S.D=Sqrt[p(1-p)/n]`

`S.D=Sqrt[0.36(0.64)/65]`

`S.D=0.05953.`

Now calculating the probability P(X≥0.4)

Adjusting the left side as we need use C.F factor as C.F=0.5/n=0.0077

Left side =0.4-0.0077=0.3923

Right side =∞

Now ,P(Z≥0.4)

=Pr(Z≥0.3923-0.36/0.0595)

=Pr(Z≥0.54)

=0.2937

Hence probability that sample proportion greater than 0.4 is 0.2937

Now calculating the probability ,P(0.30≤X≤0.4)

Similar using two tailed hypotheses we get

Now First calculate the continuity correction, as C.F=0.5/n=0.5/65=0.0077

Now adjusting the end points ,

left side=0.3-0.0077=0.2923

right side=0.4+0.0077=0.4077

Hence calculating for

P(0.2923≤X≤0.4077)

=P[(0.2923-0.36)/0.0595≤Z≤(0.4077-0.36/0.0595)]

=P[-1.14≤Z≤0.8]

=Pr(Z≤0.8)-Pr(Z≤-.1.14)

=0.7885-0.1278

=0.6607

Hence the probability that sample proportion is between in 0.30 and 0.40 is 0.6607

Now calculating the probability that sample proportion will less than 0.3

P(X≤0.3)

Using C.F factor as 0.0077 adjusting right side as ,

right side=0.3+0.0077=0.3077

P(X≤0.3077)

=Pr(Z≤0.3077-0.36/0.0595)

=Pr(Z≤-0.88)

=0.1898