Based on a recent nation-wide poll, a seller of printed advertisements estimates that 56% of all adults usually open all the mails they receive. If this is still the current rate at which adults open mail, what is the probability that, in a random sample of 1000 adults, the number who usually open all of their mails will be a) Less than 541? b) 570 or more?

Based on a recent nation-wide poll, a seller of printed advertisements estimates that 56% of all adults usually open all the mail they receive. Assuming this rate is still valid, find the probability that in a random sample of 1000 adults, the number who usually open all their mail will be:

• (a) Less than 541
• (b) 570 or more

---

Solution

The number of adults who usually open all their mail follows a Binomial Distribution because:

• Number of trials, n = 1000
• Probability of success, p = 0.56
• Probability of failure, q = 1 – p = 0.44

Since n = 1000 is very large, we use the Normal Approximation to the Binomial Distribution.

Step 1: Calculate the Mean and Standard Deviation

Mean (μ):

$$\mu = np = 1000 \times 0.56 = 560$$

Standard Deviation (σ):

$$\sigma = \sqrt{npq}$$ $$= \sqrt{1000 \times 0.56 \times 0.44}$$ $$= \sqrt{246.4}$$ $$\sigma \approx 15.70$$

Thus,

• Mean (μ) = 560
• Standard deviation (σ) ≈ 15.70

---

(a) Probability that the number is less than 541

We apply the continuity correction.

$$P(X<541)=P(X<540.5)$$

Calculate the Z-score:

$$Z=\frac{540.5-560}{15.70}$$ $$=\frac{-19.5}{15.70}$$ $$Z\approx -1.24$$

From the standard normal table,

$$P(Z<-1.24)=0.1075$$

Answer (a)

$$\boxed{P(X<541)\approx0.1075}$$

Interpretation: There is approximately a 10.75% chance that fewer than 541 adults out of 1000 will usually open all their mail.

---

(b) Probability that the number is 570 or more

Again, apply the continuity correction.

$$P(X\ge570)=P(X\ge569.5)$$

Compute the Z-score:

$$Z=\frac{569.5-560}{15.70}$$ $$=\frac{9.5}{15.70}$$ $$Z\approx0.61$$

From the standard normal table,

$$P(Z<0.61)=0.7291$$

Therefore,

$$P(Z\ge0.61)=1-0.7291$$ $$=0.2709$$

Answer (b)

$$\boxed{P(X\ge570)\approx0.2709}$$

Interpretation: There is approximately a 27.09% chance that 570 or more adults in the sample of 1000 will usually open all their mail.

---

Final Answers

Part	Probability
(a) Less than 541 adults	0.1075 (10.75%)
(b) 570 or more adults	0.2709 (27.09%)

Conclusion

Using the normal approximation to the binomial distribution, the required probabilities are:

• (a) $P(X<541)=0.1075$
• (b) $P(X\ge570)=0.2709$

These results indicate that obtaining fewer than 541 adults who usually open all their mail is relatively unlikely (about 11%), while obtaining 570 or more adults has a probability of about 27%. The normal approximation is appropriate because the sample size is large and both $np$ and $nq$ are much greater than 5, satisfying the conditions for its use.