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guapka [62]
2 years ago
14

Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the

standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weighed.Within what limits will 95% of the sample means occur?
Mathematics
1 answer:
solniwko [45]2 years ago
6 0

Answer:

(5953.52,6046.49)

Step-by-step explanation:

We are given the following in the question:

Mean, \mu = 6,000 pounds

Sample size, n = 40

Alpha, α = 0.05

Standard deviation, σ = 150 pounds

95% Confidence interval:

\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

6000 \pm 1.96(\dfrac{150}{\sqrt{40}} )\\\\ = 6000 \pm 46.4854=\\(5953.5146,6046.4854)\approx (5953.52,6046.49)

are the limits within which 95% of the sample means occur.

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A committee of 15 members is voting on a proposal. Each member casts a yea or nay vote. On a random voting basis, what is the pr
Igoryamba

Answer:

0.006% probability that the final vote count is unanimous.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they vote yes, or they vote no. The probability of a person voting yes or no is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Random voting:

So 50% of voting yes, 50% no, so p = 0.5

15 members:

This means that n = 15

What is the probability that the final vote count is unanimous?

Either all vote no(P(X = 0)) or all vote yes(P(X = 15)). So

p = P(X = 0) + P(X = 15)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.5)^{0}.(0.5)^{15} = 0.00003

P(X = 15) = C_{15,15}.(0.5)^{15}.(0.5)^{0} = 0.00003

So

p = P(X = 0) + P(X = 15) = 0.00003 + 0.00003 = 0.00006

0.006% probability that the final vote count is unanimous.

7 0
2 years ago
Jenna eats dinner at a restaurant. Her bill is $14.50 Jenna pays her bill and leaves a tip. Write an inequality to represent the
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Answer:

Step-by-step explanation:

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1 year ago
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Find the value of cosAcos2Acos3A...........cos998Acos999A where A=2π/1999
Lady bird [3.3K]
Hello,

Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.

Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)

As sin x *cos x=sin (2x) /2

P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
         *1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
 as sin(x)=-sin(2pi-x) and 2pi=1999a

sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)

So  sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".

Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)

       








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