Stats isn't really my thing, any help is appreciated
1. Assume that head sizes (circumference) of new recruits in the Canadian armed forces can be
approximated by a normal distribution with a mean of 22.8 inches and a standard deviation of 1.1
inches.
(a) What proportion of recruits have head sizes between 22 and 23 inches?
b) 5% of the head sizes exceed _____ inches?
(c) What is the probability that the average head size of the next 49 recruits will be less than
22.5 inches?
(d) Use a suitable approximation to nd the probability that at least 10 of the next 30 recruits
will have head sizes less than 23 inches?
1. Assume that head sizes (circumference) of new recruits in the Canadian armed forces can be
approximated by a normal distribution with a mean of 22.8 inches and a standard deviation of 1.1
inches.
(a) What proportion of recruits have head sizes between 22 and 23 inches?
b) 5% of the head sizes exceed _____ inches?
(c) What is the probability that the average head size of the next 49 recruits will be less than
22.5 inches?
(d) Use a suitable approximation to nd the probability that at least 10 of the next 30 recruits
will have head sizes less than 23 inches?
-
1)a)
μ = 22.8
σ = 1.1
standardize x to z = (x - μ) / σ
P( 22 < x < 23) = P[( 22 - 22.8) / 1.1 < Z < ( 23 - 22.8) / 1.1]
P( -0.7273 < Z < 0.1818) = .5714 - .2327 = .3387
(From Normal probability table)
b)
5% of head sizes exceed a standardized score z of 1.645
P( z > 1.645) = .05
z = (x - μ) / σ
1.645 = (x- 22.8) /1.1
solve for x
(1.645)(1.1) = x-22.8
x=22.8+(1.1)(1.645) = 24.6095
x=24.6
c)
Mean μ = 22.8
Standard deviation σ = 1.1
Standard error σ / √ n = 1.1 / √ 49 = 0.1571429
standardize xbar to z = (xbar - μ) / (σ / √ n )
P(xbar < 22.5) = P( z < (22.5-22.8) / 0.157143)
= P(z < -1.9091) = .0281
(from normal probability table)
d)
Find the probability of a recruit with head size < 23 inches
μ = 22.8
σ = 1.1
standardize x to z = (x - μ) / σ
P(x < 23) = P( z < (23-22.8) / 1.1)
= P(z < 0.1818) = .5714
(From Normal probability table)
μ = 22.8
σ = 1.1
standardize x to z = (x - μ) / σ
P( 22 < x < 23) = P[( 22 - 22.8) / 1.1 < Z < ( 23 - 22.8) / 1.1]
P( -0.7273 < Z < 0.1818) = .5714 - .2327 = .3387
(From Normal probability table)
b)
5% of head sizes exceed a standardized score z of 1.645
P( z > 1.645) = .05
z = (x - μ) / σ
1.645 = (x- 22.8) /1.1
solve for x
(1.645)(1.1) = x-22.8
x=22.8+(1.1)(1.645) = 24.6095
x=24.6
c)
Mean μ = 22.8
Standard deviation σ = 1.1
Standard error σ / √ n = 1.1 / √ 49 = 0.1571429
standardize xbar to z = (xbar - μ) / (σ / √ n )
P(xbar < 22.5) = P( z < (22.5-22.8) / 0.157143)
= P(z < -1.9091) = .0281
(from normal probability table)
d)
Find the probability of a recruit with head size < 23 inches
μ = 22.8
σ = 1.1
standardize x to z = (x - μ) / σ
P(x < 23) = P( z < (23-22.8) / 1.1)
= P(z < 0.1818) = .5714
(From Normal probability table)
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Let X be the random variable whose values are the recruits' head sizes. Make a new variable, Z, defined by: Z = (X - 22.8) / 1.1. Then Z is the standard normal random variable.
(a) You want the probability P(22 <= X <= 23). In terms of Z, this probability is
P((22 - 22.8)/1.1 <= Z <= (23 - 22.8)/1.1), and can be found using the probability distribution table for the standard normal r.v.: http://www.mathsisfun.com/data/standard-…
(b) 0.05 = P(Z >= ??) (Once you find the bound ?? for Z, convert it to the suitable value for X.)
(c) Assuming the head sizes of newly arrived recruits are statistically independent, denote the next 49 head sizes by X_1, X_2, ..., X_49. All these random variables are normal, with the mean and standard deviation given in the problem, and are pairwise independent. Therefore, the sought probability is:
P( (X_1 + X_2 + ... X_49) / 49 < 2.25).
Now, you'll have to look up how to figure out the prob. distribution of a sum of independent normally distributed random variables.
(a) You want the probability P(22 <= X <= 23). In terms of Z, this probability is
P((22 - 22.8)/1.1 <= Z <= (23 - 22.8)/1.1), and can be found using the probability distribution table for the standard normal r.v.: http://www.mathsisfun.com/data/standard-…
(b) 0.05 = P(Z >= ??) (Once you find the bound ?? for Z, convert it to the suitable value for X.)
(c) Assuming the head sizes of newly arrived recruits are statistically independent, denote the next 49 head sizes by X_1, X_2, ..., X_49. All these random variables are normal, with the mean and standard deviation given in the problem, and are pairwise independent. Therefore, the sought probability is:
P( (X_1 + X_2 + ... X_49) / 49 < 2.25).
Now, you'll have to look up how to figure out the prob. distribution of a sum of independent normally distributed random variables.