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ira [324]
3 years ago
8

Use technology to solve the following problem: A ferry will safely accommodate 79 tons of passenger cars. Assume that the mean w

eight of a passenger car is 1.9 tons with standard deviation 0.5 tons. If a random sample of 39 cars are loaded onto the ferry, what is the probability that the maximum safe weight will be exceeded
Mathematics
1 answer:
eimsori [14]3 years ago
6 0

Answer:

0.05155 or 5.155%

Step-by-step explanation:

We are given;

Weight of passenger cars the ferry can accommodate = 79 tons of passenger cars

Number of cars loaded = 39

Thus;

population mean weight; μ = 1.9

Since it can accommodate 79 tons of passenger cars, then their average weight should not exceed;

79/39 = 2.03 thus;

Sample mean; x¯ = 2.03

Standard deviation; σ = 0.5

Sample size; n = 39

Let's find the z-score value;

z = (x¯ - μ)/(σ/√n)

Thus;

z = (2.03 - 1.9)/(0.5/√39)

z = 0.13/0.08

z ≈ 1.63

Thus; P(z > 1.63) = 1 - P(z < 1.63)

From the z-score table attached, the p-value which is the probability is P(z < 1.63)

Thus;

P = 1 - 0.94845

P = 0.05155 or 5.155%

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16. A telemarketer makes six phone calls per hour and is able to make a sale on 30% of these contacts. During the next two hours
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Answer:

a) 23.11% probability of making exactly four sales.

b) 1.38% probability of making no sales.

c) 16.78% probability of making exactly two sales.

d) The mean number of sales in the two-hour period is 3.6.

Step-by-step explanation:

For each phone call, there are only two possible outcomes. Either a sale is made, or it is not. The probability of a sale being made in a call is independent from other calls. 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.

A telemarketer makes six phone calls per hour and is able to make a sale on 30% of these contacts. During the next two hours, find:

Six calls per hour, 2 hours. So

n = 2*6 = 12

Sale on 30% of these calls, so p = 0.3

a. The probability of making exactly four sales.

This is P(X = 4).

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

P(X = 4) = C_{12,4}.(0.3)^{4}.(0.7)^{8} = 0.2311

23.11% probability of making exactly four sales.

b. The probability of making no sales.

This is P(X = 0).

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

P(X = 0) = C_{12,0}.(0.3)^{0}.(0.7)^{12} = 0.0138

1.38% probability of making no sales.

c. The probability of making exactly two sales.

This is P(X = 2).

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

P(X = 2) = C_{12,2}.(0.3)^{2}.(0.7)^{10} = 0.1678

16.78% probability of making exactly two sales.

d. The mean number of sales in the two-hour period.

The mean of the binomia distribution is

E(X) = np

So

E(X) = 12*0.3 = 3.6

The mean number of sales in the two-hour period is 3.6.

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