A study found that 85% of sports business executives surveyed believed that sponsor development of ________ in new venues will b
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Answer:
A.
- X ~ N(6.2kg, 2.25kg²)
B. What is the median seedless watermelon weight?____ kg.
- 6.2 kg
C. What is the Z-score for a seedless watermelon weighing 8 kg?
- 1.2
D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?
- 0.2981
E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?
- 0.1411
F. The 80th percentile for the weight of seedless watermelons is _____ kg.
- 7.5 kg
Explanation:
<em>A. X ~ N(___ , ____ )</em>
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The distribution of a random variable in a sample extracted from a population that follows a normal distribution is represented by the notation:
- X ~ N(μ, σ²)
Where:
- X is the random variable
- N stands for normal distribution function
- μ is the median of the population
- σ² is the variance of the population
Here, you have:
- μ = 6.2 kg
- σ² = (1.5kg)² = 2.25 kg²
- X ~ N(6.2kg, 2.25kg²)
<em>B. What is the median seedless watermelon weight?____ kg.</em>
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The median of a random variable that follows a normal distribution is equal to the mean, thus it is 6.2 kg.
<em>C. What is the Z-score for a seedless watermelon weighing 8 kg?</em>
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The z-score is the standardized value of the random variable. It measures how far away is the variable from the mean.
It is calcuated with the formula:
Thus for X=8:
<em>D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?</em>
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You want P(X>7)
You must use the tables for the standardized normal distribution.
Find the corresponding Z-score for X = 7
You must use a table for the standardized normal distribution which gives the cumulative distribution or area under the curve of the standard normal distribution and find P(Z>0.53).
That is the area to the right of Z=0.53. The table shows 0.2981.
Thus, the probability that a randomly selected seedless watermelon will weigh more than 7 kg is 0.2981
<em>E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?</em>
<em></em>
For this case you must find the Z-scores for X=4 and X=5 and then find the area under the curve of the standardized normal distribution between those two Z-scores.
- Z(X=4) = (4 - 6.2)/1.5 ≈ -1.47
- Z(X=5) = (5 - 6.2)/1.5 = -0.8
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In the table the area to the right of Z = - 1.47 is 1 - 0.0708 = 0.9292
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And the area to the right of Z = - 0.8 is 1 - 0.2119 = 0.7881
Thus, the area in between is the difference 0.9292 - 0.7881 = 0.1411.
<em>F. The 80th percentile for the weight of seedless watermelons is _____ kg.</em>
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The 80th percentile is the weigh of the top 20% seedless watermelons: this is 80% of the weighs are below that weight.
You must find the Z-score for which the area under the curve is less than 0.80.
The area less than 0.80 is 1 less the area that is less than 0.20.
From the table, the Zscore that defines the area less than 0.20 is 0.845 (interpolating).
Thus, the 80th percentile is the X value that makes the Z-score greater than or equal to 0.845:
- Z ≥ 0.845
- (X - 6.2)/1.5 ≥ 0.845
- X ≥ 0.845 × 1.5 + 6.2
- X ≥ 7.4675
- X ≥ 7.5 kg ← answer