Ch 6 - 7, 19

6.7

Question

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Proof
We have:

σ = .51 cm, μ = 7.20 cm

f (x) = \frac{1}{.51 \sqrt{2 π}} \exp (\frac{- 1}{2} \cdot {(\frac{x - 7.2}{.51})}^{2})

P (X \geq 6.85) = \int_{- \infty}^{6.85} f (x) d x \approx .2451 \approx Φ (\frac{6.85 - 7.20}{.51})

b.
Check the probability:

P (X \leq 6.5) = \int_{- \infty}^{6.5} f (x) d x \approx .0849

So yes at least 90% of surgeons would be able to use it since $1 - .0849 = .9151$ of all surgeons surveyed have enough grip strength.

You can also say that since $ϕ (z) = .10 \Leftrightarrow z \approx - 1.28$ then $η = μ - z σ \approx 7.2 - (1.28) (.51) \approx 6.547 cm$ so then 90% of male surgeons have a grip strength greater than $η$ .

c. Do a similar process:

σ = .50 cm, μ = 6.58 cm

f (x) = \frac{1}{.50 \sqrt{2 π}} \exp (\frac{- 1}{2} \cdot {(\frac{x - 6.58}{.50})}^{2})

P (X \geq 6.85) = 1 - Φ (\frac{6.85 - 6.58}{.50}) \approx .2950

P (X \geq 6.5) = 1 - Φ (\frac{6.5 - 6.58}{.50}) \approx .5636

d. We want to calculate $η_{.95}$ for the female surgeons and compare to $μ_{male} = 7.20 cm$ :

F (η) = .95 \to η \approx 7.40 cm

Thus this value exceeds the average for the male grip strength.
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6.19

Question

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Proof
We have $μ = 2295 K$ and $σ = 477 K$ .

P (X \leq 1000) = Φ (\frac{1000 - 2295}{477}) \approx Φ (- 2.7149) \approx .0033148

P (X \geq 5000) = 1 - Φ (\frac{5000 - 2295}{477}) \approx 1 - Φ (5.6709) \approx 1 - .99999 \approx 0

Φ (\frac{z - 2295}{477}) = .10 \Rightarrow \frac{z - 2295}{477} \approx - 1.28155 \Rightarrow z \approx 1684 K

d.
First let's find the probability $p = P (X \leq 2000) = Φ (\frac{2000 - 2295}{477}) \approx Φ (- 0.6184) \approx .26815585$ . Having this, we are asking for the probability:

P (at least one) = 1 - P (none) = 1 - (1 - p)^{4} \approx .7131

We also could've used a Binomial Experiment with $n = 4$ and $p$ as described above, with $x = 0$ to get $P (none)$ .
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