In a poll about work, 70% of respondents said that their jobs were sometimes or always stressful. Ten workers are chosen at random. Round the answers to four decimal places.

To find the probability that exactly 9 out of 10 workers find their jobs stressful when 70% of respondents report this, we can use the binomial probability formula:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where:

• \(n\) is the number of trials (in this case, 10 workers),
• \(k\) is the number of successes (in this case, 9 workers),
• \(p\) is the probability of success on an individual trial (0.7 for finding the job stressful),
• \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

Substituting in the values:

• \(n = 10\)
• \(k = 9\)
• \(p = 0.7\)
• \(1 - p = 0.3\)

Now, we calculate the binomial coefficient \(\binom{10}{9}\):

\[ \binom{10}{9} = \frac{10!}{9!(10-9)!} = \frac{10!}{9! \cdot 1!} = 10 \]

Next, we calculate \(p^k\) and \((1 - p)^{n - k}\):

\[ p^k = 0.7^9 \quad \text{and} \quad (1 - p)^{n - k} = 0.3^{1} \]

Now calculate \(0.7^9\):

\[ 0.7^9 \approx 0.040353607 \]

And for \(0.3^1\):

\[ 0.3^1 = 0.3 \]

Now we can put it all together:

\[ P(X = 9) = \binom{10}{9} \cdot (0.7^9) \cdot (0.3^1) = 10 \cdot 0.040353607 \cdot 0.3 \]

Calculating:

\[ = 10 \cdot 0.040353607 \cdot 0.3 = 10 \cdot 0.0121059 \approx 0.121059 \]

Rounding to four decimal places, we get:

\[ P(X = 9) \approx 0.1211 \]

Therefore, the probability that exactly 9 of the 10 workers find their jobs stressful is 0.1211.

(b)What is the probability that more than 6 find their jobs stressful?

To calculate the probability that more than 6 out of 10 workers find their jobs stressful, we need to find the probabilities for \(X = 7\), \(X = 8\), \(X = 9\), and \(X = 10\), and then sum them up:

\[ P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \]

We'll use the binomial probability formula again:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where \(n = 10\) and \(p = 0.7\). We'll calculate each probability.

1. Calculate \(P(X = 7)\):

Using \(k = 7\):

\[ \binom{10}{7} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

Now calculate:

\[ P(X = 7) = \binom{10}{7} \cdot (0.7^7) \cdot (0.3^{3}) \]

Calculating \(0.7^7\) and \(0.3^3\):

\[ 0.7^7 \approx 0.0823543 \] \[ 0.3^3 = 0.027 \]

Putting it together:

\[ P(X = 7) = 120 \cdot 0.0823543 \cdot 0.027 \approx 0.267203649 \approx 0.2672 \]

2. Calculate \(P(X = 8)\):

Using \(k = 8\):

\[ \binom{10}{8} = \frac{10!}{8!2!} = \frac{10 \times 9}{2 \times 1} = 45 \]

Now calculate:

\[ P(X = 8) = \binom{10}{8} \cdot (0.7^8) \cdot (0.3^{2}) \]

Calculating \(0.7^8\) and \(0.3^2\):

\[ 0.7^8 \approx 0.05764801 \] \[ 0.3^2 = 0.09 \]

Putting it together:

\[ P(X = 8) = 45 \cdot 0.05764801 \cdot 0.09 \approx 0.233ensic396 \]

3. Calculate \(P(X = 9)\):

As calculated before,

\[ P(X = 9) \approx 0.1211 \]

4. Calculate \(P(X = 10)\):

Using \(k = 10\):

\[ \binom{10}{10} = 1 \]

Now calculate:

\[ P(X = 10) = \binom{10}{10} \cdot (0.7^{10}) \cdot (0.3^{0}) = 1 \cdot (0.7^{10}) \cdot 1 \]

Calculating \(0.7^{10}\):

\[ 0.7^{10} \approx 0.0282475 \]

5. Total Probability \(P(X > 6)\):

Now we can sum these probabilities:

\[ P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \]

Substituting the calculated values:

\[ P(X > 6) \approx 0.2672 + 0.2335 + 0.1211 + 0.0283 \approx 0.6501 \]

Rounding to four decimal places, we get:

\[ P(X > 6) \approx 0.6501 \]

Therefore, the probability that more than 6 of the 10 workers find their jobs stressful is 0.6501.

(c)What is the probability that fewer than 5 find their jobs stressful?

To calculate the probability that fewer than 5 out of 10 workers find their jobs stressful, we need to find the probabilities for \(X = 0\), \(X = 1\), \(X = 2\), \(X = 3\), and \(X = 4\), and then sum them up:

\[ P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \]

We'll again use the binomial probability formula:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where \(n = 10\) and \(p = 0.7\). We will calculate each probability.

1. Calculate \(P(X = 0)\):

Using \(k = 0\):

\[ \binom{10}{0} = 1 \]

Now calculate:

\[ P(X = 0) = \binom{10}{0} \cdot (0.7^0) \cdot (0.3^{10}) = 1 \cdot 1 \cdot (0.3^{10}) \]

Calculating \(0.3^{10}\):

\[ 0.3^{10} \approx 0.0000059049 \]

So:

\[ P(X = 0) \approx 0.0000059 \]

2. Calculate \(P(X = 1)\):

Using \(k = 1\):

\[ \binom{10}{1} = 10 \]

Now calculate:

\[ P(X = 1) = \binom{10}{1} \cdot (0.7^1) \cdot (0.3^{9}) = 10 \cdot 0.7 \cdot (0.3^{9}) \]

Calculating \(0.3^{9}\):

\[ 0.3^{9} \approx 0.00001937 \]

Now putting it together:

\[ P(X = 1) = 10 \cdot 0.7 \cdot 0.00001937 \approx 0.00013559 \]

3. Calculate \(P(X = 2)\):

Using \(k = 2\):

\[ \binom{10}{2} = 45 \]

Now calculate:

\[ P(X = 2) = \binom{10}{2} \cdot (0.7^2) \cdot (0.3^{8}) = 45 \cdot (0.7^2) \cdot (0.3^{8}) \]

Calculating \(0.7^{2}\) and \(0.3^{8}\):

\[ 0.7^{2} = 0.49 \] \[ 0.3^{8} \approx 0.00006451 \]

Putting it together:

\[ P(X = 2) = 45 \cdot 0.49 \cdot 0.00006451 \approx 0.001424845 \]

4. Calculate \(P(X = 3)\):

Using \(k = 3\):

\[ \binom{10}{3} = 120 \]

Now calculate:

\[ P(X = 3) = \binom{10}{3} \cdot (0.7^3) \cdot (0.3^{7}) = 120 \cdot (0.7^3) \cdot (0.3^{7}) \]

Calculating \(0.7^{3}\) and \(0.3^{7}\):

\[ 0.7^{3} \approx 0.343 \] \[ 0.3^{7} \approx 0.0002187 \]

Putting it together:

\[ P(X = 3) = 120 \cdot 0.343 \cdot 0.0002187 \approx 0.008965034 \]

5. Calculate \(P(X = 4)\):

Using \(k = 4\):

\[ \binom{10}{4} = 210 \]

Now calculate:

\[ P(X = 4) = \binom{10}{4} \cdot (0.7^4) \cdot (0.3^{6}) = 210 \cdot (0.7^4) \cdot (0.3^{6}) \]

Calculating \(0.7^{4}\) and \(0.3^{6}\):

\[ 0.7^{4} \approx 0.2401 \] \[ 0.3^{6} \approx 0.000729 \]

Putting it together:

\[ P(X = 4) = 210 \cdot 0.2401 \cdot 0.000729 \approx 0.038095211 \]

6. Total Probability \(P(X < 5)\):

Now we can sum these probabilities:

\[ P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \]

Substituting the calculated values:

\[ P(X < 5) \approx 0.0000059 + 0.00013559 + 0.001424845 + 0.008965034 + 0.038095211 \]

Calculating the total:

\[ P(X < 5) \approx 0.048631679 \]

Rounding to four decimal places, we get:

\[ P(X < 5) \approx 0.0486 \]

Therefore, the probability that fewer than 5 of the 10 workers find their jobs stressful is 0.0486.

(d)Using a cutoff of 0.05, would it be unusual if fewer than 4 of them find their jobs stressful? why or why not

To determine whether it would be unusual for fewer than 4 of the 10 workers to find their jobs stressful using a cutoff of 0.05, we need to calculate \(P(X < 4)\) and see if it is less than 0.05.

Calculation of \(P(X < 4)\)

We've already found \(P(X < 5)\) in the previous part, which includes \(P(X = 0)\), \(P(X = 1)\), \(P(X = 2)\), and \(P(X = 3)\). Now we need to find \(P(X = 4)\) as well to sum up to \(P(X < 4)\).

From previous calculations:

• \(P(X = 0) \approx 0.0000059\)
• \(P(X = 1) \approx 0.00013559\)
• \(P(X = 2) \approx 0.001424845\)
• \(P(X = 3) \approx 0.008965034\)

We already calculated \(P(X = 4)\) as:

\[ P(X = 4) \approx 0.038095211 \]

Now we can calculate \(P(X < 4)\):

\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]

Substituting the values:

\[ P(X < 4) \approx 0.0000059 + 0.00013559 + 0.001424845 + 0.008965034 \]

Calculating:

\[ P(X < 4) \approx 0.0000059 + 0.00013559 + 0.001424845 + 0.008965034 \approx 0.010531469 \]

Conclusion

Now we need to compare \(P(X < 4)\) with the cutoff of 0.05:

\[ P(X < 4) \approx 0.0105 \]

Since \(0.0105 < 0.05\), this probability is indeed less than the cutoff.

Answer:

Thus, it would be considered unusual for fewer than 4 of the 10 workers to find their jobs stressful because the probability of this occurring (0.0105) is less than the cutoff of 0.05. This indicates that such an event is likely to happen very infrequently under the given conditions.