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Alexandra [31]
3 years ago
13

If the distribution is really (5.43,0.54)

Mathematics
1 answer:
defon3 years ago
7 0

Answer:

0.7486 = 74.86% observations would be less than 5.79

Step-by-step explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:

\mu = 5.43, \sigma = 0.54

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{5.79 - 5.43}{0.54}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

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A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publica
umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

X is a random variable of Poisson distribution with mean \mu = 3 , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

  • PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
  • PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)

we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

8 0
3 years ago
Do this question and the silverfish will love you<br> also help pls
Ratling [72]

Answer:

yeah pretty much

Step-by-step explanation:

6 0
3 years ago
Ana vende productos de limpieza .Por cada $100 que venda le dan de comisión $30 .¿Cuánto ganará si vende $600?
Ulleksa [173]

Answer:

$180

Step-by-step explanation:

$600/$100 = 6

$600 es 6 veces $100.

Por vender $600 gana 6 veces mas que por vender $100.

6 * $30 = $180

Respuesta: $180

8 0
3 years ago
Can sum one help me
sleet_krkn [62]

Answer:

help with what exactly

8 0
3 years ago
Part A
AnnZ [28]

Answer:

blue green and yellow

Step-by-step explanation:

7 0
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