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Ann [662]
3 years ago
12

The average daily rainfall for the past week in the town of Hope Cove is normally distributed, with a mean rainfall of 2.1 inche

s and a standard deviation of 0.2 inches. If the distribution is normal, what percent of data lies between 1.9 inches and 2.3 inches of rainfall? a) 95% b) 99.7% c) 34% d) 68%
Mathematics
1 answer:
natita [175]3 years ago
6 0

Answer:

D

Step-by-step explanation:

We calculate the z-score for each

Mathematically;

z-score = (x-mean)/SD

z1 = (1.9-2.1)/0.2 = -1

z2 = (2.3-2.1)/0.2 = 1

So the proportion we want to calculate is;

P(-1<x<1)

We use the standard score table for this ;

P(-1<x<1) = P(x<1) -P(x<-1) = 0.68269 which is approximately 68%

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Jim Tree sells trees. The mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. Jim wants to
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Step-by-step explanation:

We know that,

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

Where,

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μ = mean = 68

σ = standard deviation = 10

Putting the values,

Z=\dfrac{84-68}{10}=1.6

The percent of his sales that were less than or equal to 84 inches, is

P(Z\leq 1.6)=0.9452=94.52\%\approx 94.5\%

Therefore, the remaining 100-94.5=5.5% were more than 84 inches.

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