Question

1)Suppose that the population distribution of the gripping strengths of industrial workers is known to have a mean of 110 and standard deviation of 10. For a random sample of 75 workers, what is the probability that the sample mean gripping strength will be

a)Between 109 and 112? 

b)Greater than112?

2)On the average, five smokers pass a certain street corners every ten minutes, what is the probability that during a given 10minutes the number of smokers passing will be 

A)6 or fewer 

B)7 or more 

C)Exactly 8

Answer 1

1)

$\mu = 110 \\ \sigma = 10 \\ n = 75 \\$

a)

$P(109<X<112) = P(X<112)-P(X<109) \\ = P(Z< \frac{112-110}{10/ \sqrt{75}}) -P(Z< \frac{109-110}{10/ \sqrt{75}}) \\ = P(Z<1.733)-P(Z< -0.866) \\ = 0.9584 -0.1932 \\ = 0.7652$

b)

$P(X>112) = 1-P(X<112) \\ = 1 -P(Z< \frac{112-110}{10/ \sqrt{75}}) \\ = 1 -P(Z<1.733) \\ = 1 -0.9584 \\ = 0.0416$

2) Assuming that the number of smokers follows Poisson distribution, if X is a Poisson random variable with parameter λ then

$P(X=x) = \frac{λ^x \times e^{-λ}}{x!} \\ λ = 5$

A) P(X≤6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)

$P(X=0) = \frac{5^0 \times e^{-5}}{0!} \\ = \frac{0.006737}{1} \\ = 0.006737 \\ P(X=1) = \frac{5^1 \times e^{-5}}{1!} \\ = \frac{0.03368}{1} \\ = 0.03368 \\ P(X=2) = \frac{5^2 \times e^{-5}}{2!} \\ = \frac{0.16844}{2} \\ = 0.08422 \\ P(X=3) = \frac{5^3 \times e^{-5}}{3!} \\ = \frac{0.84224}{6} \\ = 0.14037 \\ P(X=4) = \frac{5^4 \times e^{-5}}{4!} \\ = \frac{4.21121}{24} \\ = 0.17546 \\ P(X=5) = \frac{5^5 \times e^{-5}}{5!} \\ = \frac{21.056}{120} \\ = 0.17546 \\ P(X=6) = \frac{5^6 \times e^{-5}}{6!} \\ = \frac{105.28}{720} \\ = 0.14622 \\ P(X≤6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) \\ P(X≤6) = 0.006737 + 0.03368 + 0.08422 + 0.14037 + 0.17546 + 0.17546 + 0.14622 \\ = 0.762447$

B) P(X≥7) = 1 -P(X<7)

P(X<7) = P(≤6)

$P(X≤6) = 0.762447 \\ P(X≥7) = 1 -0.762447 \\ = 0.237553$

C) the probability that, during a given 10-minute period, the number of smokers passing the street corner will be eight is:

$P(X=8) = \frac{5^8 \times e^{-5}}{8!} \\ = \frac{2632.01}{40320} \\ = 0.065$