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lubasha [3.4K]
3 years ago
9

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 49 b

atteries and determine whether each is within specifications. The entire shipment is accepted if at most 3 batteries do not meet specifications. A shipment contains 6000 ​batteries, and 1​% of them do not meet specifications. What is the probability that this whole shipment will be​ accepted? Will almost all such shipments be​ accepted, or will many be​ rejected?
Mathematics
1 answer:
Brilliant_brown [7]3 years ago
4 0

Answer:

Step-by-step explanation:

Given that A shipment contains 6000 ​batteries, and 1​% of them do not meet specifications.

Sample size =49

If x is the number of batteries not meeting specifications then

x is binomial with n =49 and p = 0.01

Because i) each toy is independent of the other

ii) There are only two outcomes

Probability whole shipment is accepted = Prob (X≤3)

=0.9885

Hence almost all shipments would be accepted.

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Answer:

Step-by-step explanation:

46-21=25

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A money market fund advertises a simple interest rate of 88​%. Find the total amount received on an investment of ​$4000 for 15
Anna007 [38]
\bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$4000\\
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t=years\to \frac{15}{12}\to &\frac{5}{4}
\end{cases}
\\\\\\
I=4000\cdot 0.88\cdot \cfrac{5}{4}
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3 years ago
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Answer:

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3 years ago
The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =
gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is \mathbf{P(X>1100n) = 0.158655}

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy  is given as :

f_x(x)  = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0

Thus, the expected  total claim amount \mu =  1000

The variance of the total claim amount \sigma ^2  = 1000^2

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})

P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})

P(X>1100n) = P(Z> \dfrac{10*100}{1000})

P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345

\mathbf{P(X>1100n) = 0.158655}

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is \mathbf{P(X>1100n) = 0.158655}

4 0
3 years ago
A number of robbers were discussing how best to share some stolen rolls of
kupik [55]

Answer:

83

Step-by-step explanation:

Let the number of robbers be x

If they share 6 rolls each, the tota rolls will be 6x and since they have five remainder, the overall total is then 6x+5

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Equating these two

7x-8=6x+5

Arranging similar terms

7x-6x=8+5

x=13

Therefore, the robbers are 13

The rolla they have will be

6(13)+5=83

Or

7(13)-8=83

6 0
3 years ago
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