Ch 6 - 7, 19

6.7

Question

Pasted image 20241018145032.png

Proof
We have:

σ=.51 cm,μ=7.20 cmf(x)=1.512πexp(12(x7.2.51)2)

a.

P(X6.85)=6.85f(x)dx.2451Φ(6.857.20.51)

b.
Check the probability:

P(X6.5)=6.5f(x)dx.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)=.10z1.28 then η=μzσ7.2(1.28)(.51)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 cmf(x)=1.502πexp(12(x6.58.50)2)P(X6.85)=1Φ(6.856.58.50).2950P(X6.5)=1Φ(6.56.58.50).5636

d. We want to calculate η.95 for the female surgeons and compare to μmale=7.20 cm:

F(η)=.95η7.40 cm

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

6.19

Question

Pasted image 20241018153049.png
Pasted image 20241018153054.png

Proof
We have μ=2295K and σ=477K.

a.

P(X1000)=Φ(10002295477)Φ(2.7149).0033148

b.

P(X5000)=1Φ(50002295477)1Φ(5.6709)1.999990

c.

Φ(z2295477)=.10z22954771.28155z1684 K

d.
First let's find the probability p=P(X2000)=Φ(20002295477)Φ(0.6184).26815585. Having this, we are asking for the probability:

P(at least one)=1P(none)=1(1p)4.7131

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