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    The lengths of body feathers of a particular species of bird are modelled by a normal distribution. A researcher measures the lengths of a random sample of 600 body feathers from birds of this species and finds that 63 are less than 6 cm long and 155 are more than 12 cm long.
    (i) Find estimates of the mean and standard deviation of the lengths of body feathers of birds of this species. [5]

    (ii) In a random sample of 1000 body feathers from birds of this species, how many would the researcher expect to find with lengths more than 1 standard deviation from the mean? [4]

    I get the first question BUT I can Not solve the second one , can someone please explain it to me with steps and an idea of why they solved it the way they did . Any help would be great . Thank you

    The answer in the markscheme is
    needP(z<–1orz>1) = 1 – Ф1) + Ф (–1)
    = 2 – 2 × 0.8413
    = 0.3174
    number = 317
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    (Original post by Sammysammy99)
    The lengths of body feathers of a particular species of bird are modelled by a normal distribution. A researcher measures the lengths of a random sample of 600 body feathers from birds of this species and finds that 63 are less than 6 cm long and 155 are more than 12 cm long.
    (i) Find estimates of the mean and standard deviation of the lengths of body feathers of birds of this species. [5]

    (ii) In a random sample of 1000 body feathers from birds of this species, how many would the researcher expect to find with lengths more than 1 standard deviation from the mean? [4]

    I get the first question BUT I can Not solve the second one , can someone please explain it to me with steps and an idea of why they solved it the way they did . Any help would be great . Thank you

    The answer in the markscheme is
    needP(z<–1orz>1) = 1 – Ф1) + Ф (–1)
    = 2 – 2 × 0.8413
    = 0.3174
    number = 317
    If the mean was say, 65 and the standard deviation was 5, then you'd be looking at P(X>70) and and X<60, where X is a single body feather. Why? Because below 60 and above 70, the values are more than 1 standard deviation from the mean. It's confusing but you just calculate the values like I just did, then convert to z values and voila

    Unless it's the 1-.. bit that's confusing you?
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    (Original post by Sammysammy99)
    (ii) In a random sample of 1000 body feathers from birds of this species, how many would the researcher expect to find with lengths more than 1 standard deviation from the mean? [4]

    The answer in the markscheme is
    needP(z<–1orz>1) = 1 – Ф1) + Ф (–1)
    = 2 – 2 × 0.8413
    = 0.3174
    number = 317
    To really understand this question, you need to remember what a normal distribution really is.
    The bell curve (graph showing the normal distribution) is symmetrical. Half of the data will fall to the left of the mean; half will fall to the right.

    Personally, with statistics - it's best to read about the topic, especially different distributions to really understand them inside out.

    • 68% of the data falls within one standard deviation of the mean.
    • 95% of the data falls within two standard deviations of the mean.
    • 99.7% of the data falls within three standard deviations of the mean.

    How do you know this?

    This is what will help answer your question and help you with general knowledge of the normal distribution too.

    The number of standard deviations is represented by the value of z.
    For example, the standardised normal distribution Z = (X - μ)/σ.
    where Z ~ N (0, 1^2).

    So a mean of 0 and a standard deviation of 1 and therefore a variance of 1^2 = 1.

    So, when approximating using the normal distribution, the value of Z really determines the standard deviations that the X value is above or below the mean.

    For example, if you rearrange that equation: Zσ = X - μ --> Zσ + μ. Since σ = 1, Z represents the standard deviations that the value of X is away from the mean. X = Z + μ. If Z = -1, the value of X is 1 standard deviation below the mean, as it is left of the curve/mean. If the value of Z is 1, the value of X is 1 standard deviation above the mean, as it is to the right of the curve/mean.

    Coming back to "68% of the data falls within one standard deviation of the mean".

    Using the formula booklet, at least the one provided by Edexcel. The probability of the data falling within one standard deviation of the mean, means that it is either 1 standard deviation below or above the mean.
    Therefore, Z ≥ -1 or Z ≤ 1.
    So, P ( -1 ≤ Z ≤ 1) = P (-1 < Z < 1).
    You know that P (Z ≤ 1) = P (Z < 1) from the tables are 0.8413. So, the probably of (0 < Z < 1) is P(Z < 1) - P (Z < 0) = 0.8413 - 0.5 = 0.3413.
    Since the curve is symmetrical, P (-1 < Z < 0) = P(0 < Z < 1). Therefore, P (-1 < Z < 1) = P (-1 < Z < 0) * 2 OR P (0 < Z < 1) * 2 = 0.3413 * 2 = 0.6826. Therefore, the percentage of data that falls within one standard deviation of the mean is 0.6826 * 100% = 68.26%.

    So, how many would the researcher expect to find with lengths more than 1 standard deviation from the mean?

    It is the opposite of this. 1 - 0.6826 = 0.3174. If I said that the probability of the data falling within one standard deviation of the mean is 0.6826. It means that the rest of the data falls outside or more than 1 standard deviations of the mean. Therefore, 1.2, 2, 3, etc. standard deviations of the mean. Alternatively, you know Z = standard deviation from the mean. So, more than 1 standard deviation from the mean = more than 1 standard deviation below and above the mean so Z < -1 or Z > 1.


    Working out the probabilities gives you the answer. Since there are 1000 people, the number of people will be probability * sample size = 0.3174 * 1000 = 317.4 = 317 (nearest integer) because you can't have .4 people.
 
 
 
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