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weqwewe [10]
2 years ago
10

A population of insect weights is normal with a mean of μ = 16. If the probability of a weight falling between 10 and 22 is 95.4

4%, what is the standard deviation for this distribution
a. 1.5
b. 3
c. 6
d. 12
Mathematics
1 answer:
mrs_skeptik [129]2 years ago
8 0

Answer:

6

Step-by-step explanation:

This is calculated as using z score

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Hence,

We find the z score of 95.44% confidence interval

95.44% = 1.9991

Hence.

1.9991 = ( X = 22) - (X = 10)

1.9991 = (22 - 16/ σ) - (10 - 16/ σ)

1.9991 = (6/σ - (-6/σ)

1.9991 = ( 6/σ + 6/σ)

1.9991 = (6 + 6/σ)

1.9991 = 12/σ

Cross Multiply

= 1.9991 × σ = 12

σ = 12/1.9991

= 6.00270121555

Approximately 6

Hence, the standard deviation is 6

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Part (2): getting the initial value:
The initial value is defined as the output when the input is zero. Therefore, it is the value of the "y" when the value of the "x" is zero.
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Hope this helps :)
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