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Ket [755]
3 years ago
9

Suppose the test scores on an exam show a normal distribution with a mean of 82 and a standard deviation of

Mathematics
1 answer:
Zanzabum3 years ago
5 0

Answer:

a) Between 74 and 90

b) 81.86% of the scores are between 78 and 90

Step-by-step explanation:

a) According to the Empirical Rule, in a normal distribution, 95% of the data are within 2 standard deviations of the mean. Therefore, the range that 95% of the scores fall in is between 82-4(2)=74 and 82+4(2)=90.

b) The percent of scores that are between 78 and 90 is: normalcdf(lower,upper,μ,σ) = normalcdf(78,90,82,4) = 0.8185946784 ≈ 0.8186, or about 81.86% of the scores.

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Aleksandr-060686 [28]
You must think h,p and q are just numbers and m is unknown then solve it for m. The answer is A
3 0
3 years ago
The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed.
tatuchka [14]

Answer:

a)Null hypothesis:- H₀: μ> 500

  Alternative hypothesis:-H₁ : μ< 500

b) (5211.05 , 5411.7)

95% lower confidence bound on the mean.

c) The test of hypothesis t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Step-by-step explanation:

<u>Step :-1</u>

Given  a random sample of 15 devices is selected in the laboratory.

size of the small sample 'n' = 15

An average life of 5311.4 hours and a sample standard deviation of 220.7 hours.

Average of sample mean (x⁻) =  5311.4 hours

sample standard deviation (S) = 220.7 hours.

<u>Step :- 2</u>

<u>a) Null hypothesis</u>:- H₀: μ> 500

<u>Alternative hypothesis</u>:-H₁ : μ< 500

<u>Level of significance</u> :- α = 0.95 or 0.05

b) The test statistic

t = \frac{x^{-} - mean}{\frac{S}{\sqrt{n-1} } }

t = \frac{5311.4 - 500}{\frac{220.7}{\sqrt{15-1} } }

t = 5.826

The degrees of freedom γ= n-1 = 15-1 =14

tabulated value t =1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

calculated value t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Null hypothesis is rejected at  95% confidence on the mean.

C) <u>The 95% of confidence limits </u>

(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-} + t_{0.05}\frac{S}{\sqrt{n} } )

substitute values and simplification , we get

(5311.4 - 1.761 \frac{220.7}{\sqrt{15} } ,5311.4 +1.761\frac{220.7}{\sqrt{15} } )

(5211.05 , 5411.7)

95% lower confidence bound on the mean.

5 0
3 years ago
Kim places £500 in a bank account the pays 3% simple interest each year
IgorLugansk [536]
She will gain 3% each year and after one year she will gain £15 leaving her total at £515
8 0
3 years ago
Which expression is equivalent to n2 + 26n + 88 for all values of n?
Lelechka [254]

Answer:

Option B is correct.

The expression which is equivalent to  n^2+26n+88  is; (n+22)(n+4)

Explanation:

Given the equation: n^2+26n+88 for all values of n.

The quadratic equation in this form; ax^2+bx+c =0

First find two numbers that multiply to give ac and add to give b.

From the given equation;

a =1 , b = 26 and c =88

then,

the two number that multiply to give ac =88 is, 22 or 4

and they add up to give b=26 (i.e 22+4)

Now, rewrite the middle term i.e 26n with 22n and 4n , we have

n^2+22n+4n+88=0

Now, factor the first two terms and last two terms,

n(n+22)+4(n+22)

we see that (n+22) is common to both terms so, we have;

(n+22)(n+4)

Therefore, the expression (n+22)(n+4) is equivalent to  n^2+26n+88

Check:

(n+4)(n+22) = n\cdot n+ 22n+4n+88=n^2+26n+88   [ True]



5 0
3 years ago
Read 2 more answers
1) A boat traveled 448 miles downstream and back. The trip downstream took 14 hours. The trip
Serhud [2]

Answer:B

I think we are in the same class lol

Step-by-step explanation:448 divided by 14= 32

448 divided by 16= 28

B=speed of boat in water

Y= Speed of current

B+y=32

B-y=28

32 + 28 =60

60 divided by 2 = 30

32 -30 =2

B=30

Y=2

Add my discord if you need more help Snapple#0141

6 0
3 years ago
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