Surrey of women (20-29) mean height was 63.8 stdiv 2.75. What height represents

Question

Surrey of women (20-29) mean height was 63.8 stdiv 2.75. What height represents 95th ? What height represents the first quartile? Time spent for transplant people ages 35-49 can be approx by normal distribution m = 1082 theta = 208.6 A) What waiting time represent the 98th of B) What waiting time represent the first quartile? Researcher wishes to estimate, with 99% confidence, the population proportion of adults who chocolate B their favorite cream. Her estimate must be 2 degree 6 of the population proportion. A) no preliminary estimate is available. Find minimum sample needed B) Find the minimum sample 3.6 needed, using a prior study, that fund 40 degree of the respondents said their flavor of ice cream is chocolate.

Explanation / Answer

Solution

Q5

Let X = height of women. Then, X ~ N(µ, 2).

We are given µ = 63.8 and = 2.75.

Part (1)

95th percentile

Let T be the 95th percentile. Then, by definition, P(X T) = 0.95 [i.e., 95%]

Now,

If X ~ N(µ, 2), then, Z = (X - µ)/ ~ N(0, 1) and hence

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }]

So, P(X T) = 0.95 => P[Z {(T – 63.8)/2.75}] = 0.95

=> {(T – 63.8)/2.75} = 95th percentile of N(0, 1).

Using Excel Function on Normal Distribution, we get {(T – 63.8)/2.75}= 1.645 or

T = (1.645 x 2.75) + 63.8 = 68.32375 or 68.32 ANSWER 1

Part (2)

Identical to the above analysis, if Q is the first quartile, then we should have:

P(X Q) = 0.25 [1/4 th of100%]

=> P[Z {(Q – 63.8)/2.75}] = 0.25

=> {(Q – 63.8)/2.75} = 25th percentile of N(0, 1).

Using Excel Function on Normal Distribution, we get {(Q – 63.8)/2.75}= - 0.67449 or

Q = (- 0.67449 x 2.75) + 63.8 = 61.94515 or 61.95 ANSWER 2

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