Question

(a) The probability that a man aged 60 will live to be 70 is 0.65. What is the probability that out of 10 men, now aged60 (i) exactly 9 will live to be 70 (ii) at most 9 will live to be 70, and (iii) at least 7 will live to be 70?

(b) In an examination taken by 500 candidates, the average and S.D of marks obtained are 40% and 10% respectively. Assuming normal distribution, find (i) how many have scored above 60%,(ii) how many will pass if 50% is fixed as the minimum marks for passing, (iii) how many will pass if 40% is fixed as the minimum marks for passing, and (iv) what should be the minimum percentage of marks for passing so that 350 candidates pass.

Answer 1

Answer on Question #45609 – Math – Statistics and Probability

(a) The probability that a man aged 60 will live to be 70 is 0.65. What is the probability that out of 10 men, now aged 60 (i) exactly 9 will live to be 70 (ii) at most 9 will live to be 70, and (iii) at least 7 will live to be 70?

(b) In an examination taken by 500 candidates, the average and S.D of marks obtained are 40% and 10% respectively. Assuming normal distribution, find (i) how many have scored above 60%, (ii) how many will pass if 50% is fixed as the minimum marks for passing, (iii) how many will pass if 40% is fixed as the minimum marks for passing, and (iv) what should be the minimum percentage of marks for passing so that 350 candidates pass.

Solution

(a) The probability that a man aged 60 will live to be 70 is $p = 0.65$.

$$q = 1 - p = 1 - 0.65 = 0.35.$$

Number of men is $n = 10$.

(i) Probability that exactly 9 will live to be 70 is

$$P(9) = \binom{10}{9} p^9 q = \frac{10!}{9! 1!} 0.65^9 \cdot 0.35 = 10 \cdot 0.65^9 \cdot 0.35 = 0.0725.$$

(ii) Probability that at most 9 will live to be 70 is

$$P(k \leq 9) = 1 - P(10) = 1 - \binom{10}{10} p^{10} q^0 = 1 - \frac{10!}{10! 0!} 0.65^{10} \cdot 1 = 1 - 0.0135 = 0.9865.$$

(iii) Probability that at least 7 men will live to 70 is

$$P(7 \text{ or } 8 \text{ or } 9 \text{ or } 10) = \binom{10}{7} p^7 q^3 + \binom{10}{8} p^8 q^2 + \binom{10}{9} p^9 q + \binom{10}{10} p^{10} q^0.$$

We already know $P(9)$ and $P(10)$. So

$$P(7) = \binom{10}{7} p^7 q^3 = \frac{10!}{7! 3!} 0.65^7 \cdot 0.35^3 = 120 \cdot 0.65^7 \cdot 0.35^3 = 0.2522,$$ $$P(8) = \binom{10}{8} p^8 q^2 = \frac{10!}{8! 2!} 0.65^8 \cdot 0.35^2 = 45 \cdot 0.65^8 \cdot 0.35^2 = 0.1756.$$ $$P(7 \text{ or } 8 \text{ or } 9 \text{ or } 10) = 0.2522 + 0.1756 + 0.0725 + 0.0135 = 0.5138.$$

(b) The parameters $\mu$ and $\sigma$ are the mean and standard deviation, respectively, and define the normal distribution. The given values $N = 500$ (number of candidates), Mean $\mu = 40$ and $\sigma = 10$.

(i) To find how many have scored above 60% we apply the following formula

$$z = \frac{x - \mu}{\sigma}.$$

We substitute the given values.

$$z = \frac{60 - 40}{10} = 2.$$

We use the table which shows the area from 0 to Z. In our case

$$P(x \geq 60) = P(z \geq 2) = P(0 < z < \infty) - P(0 < z < 2)$$

will be equal to

$$0.5 - (\text{Area between 0 and 2}) = 0.5 - 0.4772 = 0.0228.$$

According to the condition of the task, we need to find the number of candidates scored above 60%.

The required number of candidates who scored more than 60% marks = 500 · 0.0228 ≈ 11 (approximately).

(ii)

$$z = \frac{50 - 40}{10} = 1.$$

We use the table which shows the area from 0 to Z. In our problem

$P(x \geq 50) = P(z \geq 1) = 0.5 - P(0 < z \leq 1)$ will be equal to 0.5 - (Area between 0 and 1) = 0.5 - 0,3413 = 0.1587$.

Now we can find the number of candidates which will pass if 50% is fixed as the minimum marks for passing.

The required number of candidates is equal to 500 · 0.1587 = 79.35 ≈ 79 (approximately).

(iii)

$$z = \frac{40 - 40}{10} = 0.$$

We obtained 0. If the Z score of x is zero, then the value of x is equal to the mean. In our case we have mean equal to 40%. So we can write that the number of candidates which will pass if 40% is fixed as the minimum marks for passing will be equal to 500 · 0.4 = 200 candidates.

(iv) If it is known that 350 candidates are to pass then the probability of passing will be equal $\frac{350}{500} = 0.7$.

If we have $z_1$ is the minimum cut of mark then we can note the following

$$P(z \geq z_1) = 0.7 = 0.5 + 0.2 = P(0 < z < \infty) + P(0 < z < 0.525) = P(z \geq -0.525).$$

From the formula above we can write the following $z_1 = -0.525$.

Then we can substitute into the formula

$$-0.525 = \frac{x - 40}{10} \rightarrow x = 0.3475 \approx 35\%.$$

Finally we can note that 35% minimum pass marks could enable 350 candidates to pass out of 500 candidates.

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