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RUDIKE [14]
4 years ago
13

Use the given values of n and p to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ. Round your answer to

the nearest hundredth unless otherwise noted. n = 1042, p = 0.80
Mathematics
1 answer:
fenix001 [56]4 years ago
8 0

Given Information:  

number of trials = n = 1042

Probability of success = p = 0.80

Required Information:  

Maximum usual value = μ + 2σ = ?

Minimum usual value = μ - 2σ = ?

Answer:

Maximum usual value = 859.51

Minimum usual value = 807.78

Step-by-step explanation:

In a binomial distribution, the mean μ is given by

μ = np

μ = 1042*0.80

μ = 833.6

The standard deviation is given by

σ = √np(1 - p)

σ = √1042*0.80(1 - 0.80)

σ = √833.6(0.20)

σ = 12.91

The Maximum and Minimum usual values are

μ + 2σ = 833.6 + 2*12.91

μ + 2σ = 833.6 + 25.82

μ + 2σ = 859.51

μ - 2σ = 833.6 - 25.82

μ - 2σ = 807.78

Therefore, the minimum usual value is 807.78 and maximum usual value is 859.51

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For what value of a does 9=(1/27)^a+3
wlad13 [49]
Given: <span>9=(1/27)^(a+3)

[Note:  Please review rules of PEMDAS.  
The question posted as is means
</span><span>9=[(1/27)^a]+3 which is probably not what you mean.  It has been interpreted with (a+3) as the exponent, please check.]
</span><span>
Need to find value of a that satisfies above.

We will first isolate a in order to solve for its value.
</span>9=(1/27)^(a+3)
Switch unknown to the left
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