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DIA [1.3K]
2 years ago
12

Adult panda weights are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds. The largest pandas

weigh over 250 pounds. Using the empirical rule, approximately what percent of the adult pandas weigh over 250 pounds?
16%
32%
47.5%
95%
Mathematics
2 answers:
joja [24]2 years ago
4 0

Answer:

16%

Step-by-step explanation:

THERE

alukav5142 [94]2 years ago
3 0

Answer: 16%

Step-by-step explanation:

Given: Mean : \mu = 200\text{ pounds}

Standard deviation : \sigma=50\text{ pounds}

The formula for z score is given by :-

z=\dfrac{x-\mu}{\sigma}

Now, for x=250

z=\dfrac{250-200}{50}=1

The P-value = P(z>1)=1-P(z

=1-0.8413=0.1587\approx16\%

Hence, the percent of the adult pandas weigh over 250 pounds is about 16%.

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Ribbon is only sold by the spool. A spool contains 25 yards of ribbon. Shannon needs 58 yards of ribbon to wrap all the presents
ella [17]

Answer:

\boxed{\text{3 spools}}

Step-by-step explanation:

\text{1 spool =25 yd}\\\text{Spools needed } = \text{58 yd} \times \dfrac{\text{1 spool}}{\text{25 yd}} = \text{2.32 spools}\\\\\text{However, Shannon can't buy part of a spool. She must $\textbf{round up}$ to the next integer.}\\\text{Shannon must buy $\boxed{\textbf{3 spools}}$ of ribbon}

6 0
2 years ago
Read 2 more answers
The amount of time it takes for a student to complete a statistics quiz is uniformly distributed (or, given by a random variable
topjm [15]

Answer:

(A) 0.15625

(B) 0.1875

(C) Can't be computed

Step-by-step explanation:

We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let X = Amount of time taken by student to complete a statistics quiz

So,   X ~ U(32 , 64)

The PDF of uniform distribution is given by;

    f(X) = \frac{1}{b-a} ,  a < X < b      where a = 32 and b = 64

The CDF of Uniform distribution is P(X <= x) = \frac{x-a}{b-a}

(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)

   P(X > 59) = 1 - P(X <= 59) = 1 - \frac{x-a}{b-a} = 1 - \frac{59-32}{64-32} = 1-\frac{27}{32} = 0.15625

(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43)  = P(X <= 43) - P(X < 37)

    P(X <= 43) = \frac{43-32}{64-32} = \frac{11}{32} = 0.34375

    P(X < 37) = \frac{37-32}{64-32} = \frac{5}{32} = 0.15625

    P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875

(C) Probability that student complete the quiz in exactly 44.74 minutes

     = P(X = 44.74)

The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.

3 0
2 years ago
Aarti bought a square shaped table cloth for her home the side of the table cloth measures 2 1/3m what is the area of the table
PSYCHO15rus [73]

Answer:

Area of table cloth =  49/9 m² or 5.44 m²

Step-by-step explanation:

Given:

Side of table cloth  = 2\frac{1}{3}m = 7/3 m

Shape of cloth is square

Find:

Area of table cloth

Computation:

Area of square = side²

So,

Area of table cloth = side²

Area of table cloth = (7/3)²

Area of table cloth =  49/9 m² or 5.44 m²

3 0
1 year ago
A small city has three automobile dealerships: a GM dealer selling Chevrolets and Buicks; a Ford dealer selling Fords and Lincol
9966 [12]

Answer:

Event A = { Chevrolet , Buick }

Event B = { Ford , Lincoln }

Event C = { Toyota }

Step-by-step explanation:

- Mutually exclusive events are such that their probability of coming true simultaneously is zero. If we consider set notations we could say.

                             P (A & B) = P (B & C) = P (A & C) = 0

- In our case these events A,B, and C can be defined as:

Answer:

Event A = { Chevrolet , Buick }

Event B = { Ford , Lincoln }

Event C = { Toyota }

4 0
2 years ago
Last year your professor wanted to study the pattern in class grades. The following table contains a sample of grades collected
Maslowich

Answer:

mean = 78.4

median = 77.5

mode = 75

This is Right - skewed (positive skewness) distribution

Step-by-step explanation:

<u>Mean:-</u>

The mean (average) is found by adding all of the numbers together and dividing by the number of items and it is denoted by x⁻

                                                                                                                           mean =  \frac{64+80+75+98+75}{5}

mean (x⁻ ) = 78.4

The mean of the given data = 78.4

<u>Median:</u>

The median is found by ordering the set from lowest to highest and finding the exact middle.

64 ,75, 80, 98

The middle term of the given data set = \frac{75+80}{2} =77.5

<u>Mode :</u>

The mode is the most common repeated number in a data set.

64 ,75, 75, 80, 98

in data the most common number = 75

<u>Conclusion</u>:-

mean = 78.4

median = 77.5

mode = 75

This is Right - skewed (positive skewness) distribution

   

8 0
2 years ago
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