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3 years ago

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The weekly wages of employees of Volta gold are normally distributed about a mean of $1250 with a variance of $120. Find the pro

bability of an employee having a weekly wage lying: i. Between $1320 and $970. ii. Under $1400. iii. Over $1290

1 answer:

expeople1 [14]3 years ago

7 0

Answer:

0.7102

0.8943

0.3696

Step-by-step explanation:

Given :

Mean , μ = 1250

Standard y, σ = 120

A.) Between $1320 and $970

P(Z < (x - μ) /σ) - P(Z < (x - μ) / σ))

P(Z < (1320 - 1250) /120) - P(Z < (970 - 1250) / 120))

P(Z < 0.5833) - P(Z < - 2.333)

Using the standard normal table or Z probability calculator :

0.7200 - 0.0098 = 0.7102

=0.71023

B. Under 1400

x = 1400

P(Z < (1400 - 1250) /120

P(Z < 1.25) = 0.8943

C.)

Over 1290

P(Z > Z)

P(Z > (1290 - 1250) /120 = 0.3333

P(Z > z) = 1 - P(Z < 0.3333) =

P(Z < 0.3333) = 0.6304

P(Z > 0.3333) = 1 - 0.6304 = 0.3696

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A local amusement park charges $21.50 per daily adult ticket and $14.75 per daily child's ticket. A group of 12 people paid $204

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Answer:

$x+y=12$

$21.50x+14.75y=204$

Step-by-step explanation:

Let x be the number of adult tickets and y be the number of children's tickets.

We are given that there are total 12 people. Therefore, we can set:

$x+y=12$

Moreover, we are given that an adult ticket costs $21.50 and a child's ticket costs $14.75, therefore, cost of x adult tickets will be $21.50x$ and cost of y children's ticket will be $14.75y$ . We can form the second equation by setting the total cost of tickets as:

$21.50x+14.75y=204$

Therefore, the required system of equations that could be used to find x and y will be:

$x+y=12$

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. Express the following as a percent: 5 out of 30. Round to the nearest whole number

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A geologist has collected 5 specimens of basaltic rock and 7 specimens of granite. The geologist instructs a laboratory assistan

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The rocks are chosen without replacement, which means that the hypergeometric distribution is used to solve this question. First we get the parameters, and then we answer the questions. From this, we get that:

$E(X) = 5.25, Var(X) = 0.5966$
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Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

The mean and the variance are:

$\mu = \frac{nk}{N}$

$\sigma^2 = \frac{nk(N-k)(N-n)}{N^2(N-1)}$

We have that:

5 + 7 = 12 rocks, which means that $N = 12$

9 are chosen, which means that $n = 9$

7 are granite, which means that $k = 7$

Question a:

$E(X) = \mu = \frac{9\times7}{12} = 5.25$

$Var(X) = \sigma^2 = \frac{9\times7(12-7)(12-9)}{12^2(12-1)} = 0.5966$

Thus:

$E(X) = 5.25, Var(X) = 0.5966$

Question b:

Since there are only 5 specimens of basaltic rock, at least 9 - 5 = 4 specimens of granite are needed, which means that:

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,12,9,7) = \frac{C_{7,4}*C_{5,5}}{C_{12,9}} = 0.1591$

$P(X = 5) = h(5,12,9,7) = \frac{C_{7,5}*C_{5,4}}{C_{12,9}} = 0.4773$

$P(X = 6) = h(6,12,9,7) = \frac{C_{7,6}*C_{5,3}}{C_{12,9}} = 0.3181$

Thus

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.1591 + 0.4773 + 0.3181 = 0.9545$

So P(X < 6) = 0.9545.

Question c:

5 of basaltic and 4 of granite: 0.1591 probability.

7 of granite is P(X = 7), in which

$P(X = 7) = h(7,12,9,7) = \frac{C_{7,7}*C_{5,2}}{C_{12,9}} = 0.0455$

0.1591 + 0.0455 = 0.2046, thus:

P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

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