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Olenka [21]
3 years ago
11

A researcher is interested in the lengths of Salvelinus fontinalis (brook trout), which are known to be approximately Normally d

istributed with mean 80 centimeters and standard deviation 5 centimeters. To help preserve brook trout populations, some regulatory standards need to be set limiting the size of fish that can be caught.
What proportion of fish are larger than 86 centimeters in length

a. 0.0228
b. 0.0548
c. 0.1151
d. 0.8849
Mathematics
1 answer:
kompoz [17]3 years ago
5 0

Answer:

c. 0.1151

Step-by-step explanation:

d) the probability that, on average, fish are larger than 86 centimeters in length.The area under part of a normal probability curve is  directly proportional to probability and the value is calculated as

z = (x₁−x) /σ

where z = propability of normal curve

x₁ = variate mean = 86cm

x = mean of 80cm

σ = standard deviation = 5cm

applying the formula,

z= (86-80)/5

z = 6/5 =1.2

Using a table of partial areas beneath the standardized normal curve (see Table of normal curve, a z-value of 1.2 corresponds to an area of 0.3849 between the mean value. but, because the standard curve has 0.5, then will minus 0.3849 from 0.5= 0.5 - 0.3849 = 0.1151

Thus the probability of a fish are larger than 86 centimeters in length is 0.1151

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Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature
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Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

Since is a one left tailed test the p value would be:  

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0
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Can anyone help solve question3 the (ii) and the question2 match the following
Trava [24]

Answer:

2.

i) (c)

ii) (d)

iii) (b)

iv) (a)

3.

ii)  80°

Step-by-step explanation:

2.

i) 2x+x+3x=180°

x=30°

ii) x+110=180

x=70

iii) x+2x+30°=180

x=50°

iv) 2x+15°+45°+x=180

3x+60°=180

x=40°

3.

ii) 2x+5+25=180°

2x+30°=180°

x=75°

the measure:  x+5°= 75°+5°=80°

3 0
3 years ago
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