Answer:
75 plants will be taller
Step-by-step explanation:
Given mean = 64 and standard deviation = 3.5
To get normal curve , we find one , two and three standard deviation from the mean.
To get one standard deviation we add and subtract 3.5 from mean
-1 SD =60.5 , +1 SD = 67.5
-2 SD =57 , +2 SD = 71
-3 SD = 53.5 , +3 SD = 74.5
The graph is attached below with percentages
(b) from the graph we can see that, 2.5% are after 71
that means 2.5% plants are taller than 71 inches
3000* 2.5% = 3000*0.025 = 75
75 plants will be taller
Empirical rule helps getting values at 1,2, and 3 standard deviation from mean. For given case, they are:
B) Approx 75 plants area taller than 71 inches height.
According to the empirical rule, also known as 68-95-99.7 rule, the percentage of values that lie within an interval with 68%, 95% and 99.7% of the values lies within one, two or three standard deviations of the mean of the considered normal distribution.
[tex]P(\mu - \sigma < X < \mu + \sigma) = 68\%\\P(\mu - 2\sigma < X < \mu + 2\sigma) = 95\%\\P(\mu - 3\sigma < X < \mu + 3\sigma) = 99.7\%[/tex]
where we had
[tex]X \sim N(\mu, \sigma)[/tex]
where mean of distribution of X is [tex]\mu[/tex] and standard deviation from mean of distribution of X is [tex]\sigma[/tex]
For the given case, let we have a random variable X track the height of the considered field's sunflowers. Then, by the given data, we have:
[tex]X \sim N(64, 3.5)[/tex]
Thus, the range of heights with them being at one, two and three standard deviation from the mean of the distribution is
[tex]P(\mu - \sigma < X < \mu + \sigma) = 68\%\\\\P(65-3.5 = 61.5 < X < 65 + 3.5 = 68.5) = 68\%\\\\\P(\mu - 2\sigma < X < \mu + 2\sigma) = 95\%\\\\P(65-7 = 58 < X < 65 + 7=72) = 95\%\\\\P(\mu - 3\sigma < X < \mu + 3\sigma) = 99.7\%\\\\P(65-10.5 = 54.5 < X < 65 + 10.5 = 75.5) = 99.7\%\\[/tex]
Values 61.5 and 68.5 inches are at one standard deviation from the mean of X
Values 58 and 72 are at 2 standard deviation from the mean of X
Values 54.5 and 75.5 are at 3 standard deviation from the mean of X.
B: For the given case, we need to find out the percentage of the heights recorded which are higher than 71 inches, that means, approximately 72 or above. That probability can be written as:
[tex]P(X > 71)[/tex]
Since we've got X having symmetrical distribution, thus,
[tex]P(\mu - a < X < \mu + a) = 1 - [P(X < \mu - a) + P(X > \mu + a)] = 1 - 2P(X\geq \mu + a)\\P(58 < X < 72) = 95\%=0.95\\0.95 = 1-2P(X \geq72)\\P(X \geq 72) = 0.05/2 = 0.025[/tex]
Thus, we've got [tex]P(X > 71) \approx P(X \geq 72) = 0.025[/tex] = 2.5%
Since there are 3000 plants given, its 2.5% is : 3000 times 2.5/100 = 75
Approx 75 plants area taller than 71 inches height.
Thus,
For given case, the needed probability and values are:
B) Approx 75 plants area taller than 71 inches height.
Learn more about empirical rule of probability here:
https://brainly.com/question/13676793
