**(13)**

**(a)** Mean =

We need to the total weights of the passengers to be 3500 lb (less than or equal to 3500 lb)b. Since there are 25 passengers the mean will

= =

**(b)** Let weights of the men

According to the central limit, if are independent and identically distributed variables, the mean distribution of these variables will follow normal distribution with parameters and

We need to find the probability that the mean weight is more than 140 lb.

=

= ...........Using normal dist tables, excel func *'normsdist(6.28)'*

*We can get 0.99999 but I am approximating to 1.*

**(c)** Now instead of n= 25 we have n = 20. So we have the following distribution

=

**Maximum mean weight**

Total weight should not exceed 3500 lb

Max mean weight of 20 men =

= **lb**

**(d)** Since the probability that maximum mean weight 175 will be exceeded is 0.9458, which is too high, thats means the probability that total load will also have p=0.9458 of exceeding 3500 lb.

**Therefore 20 passengers arrangements is not suitable since probability of overloading is high.**

**(14)** Let Weight of coins

The coin will be accepted only if it weighs between 5.550g - 5.790

Probability of coin being accepted

=

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140 1Ь

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X, Νίμ = 189, σ = 39)

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X25 • Nụ = 189, s/s */

39 X35 ~ N(189,

2 – score = I - 189 39/25

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= P(Z > -6.282)

P(Z < 6.282)

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= P(Z < 5.790 – 5.670 0.069 ) - P(Z < 5,550 - 5.670 0.062

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