username1946
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Can someone help with d? I'll post my working

A health centre claims that the time a doctor spends with a patient can be modelled by a
normal distribution with a mean of 10 minutes and a standard deviation of 4 minutes.

(a) Using this model, find the probability that the time spent with a randomly selected
patient is more than 15 minutes.
(1)

Some patients complain that the mean time the doctor spends with a patient is more than
10 minutes.

The receptionist takes a random sample of 20 patients and finds that the mean time the
doctor spends with a patient is 11.5 minutes.

(b) Stating your hypotheses clearly and using a 5% significance level, test whether or
not there is evidence to support the patients’ complaint.
(4)

The health centre also claims that the time a dentist spends with a patient during
a routine appointment, T minutes, can be modelled by the normal distribution
where T ~ N(5, 3.52)

(c) Using this model,

(i) find the probability that a routine appointment with the dentist takes less than 2
minutes
(1)
(ii) find P (T < 2 | T > 0)
(3)
(iii) hence explain why this normal distribution may not be a good model for T.
(1)

The dentist believes that she cannot complete a routine appointment in less than 2 minutes.

She suggests that the health centre should use a refined model only including values
of T > 2

(d) Find the median time for a routine appointment using this new model, giving your
answer correct to one decimal place.
(5)
(Total for Question 5 is 15 marks)
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mqb2766
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Can you pls read the forum sticky about not posting just questions without any working/attempt/clear description of your problem. It's at the top of the forum.
https://www.thestudentroom.co.uk/sho....php?t=4919248

This is the 5th or 6th one today. Also, don't reply via PMs. Reply to the thread pls.
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username1946
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P(T>2) = 0.8043

median is (1-0.8043)/2 = 0.0979

Area = 0.90215

Standardising:
z = ??? not a clue
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username1946
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(Original post by mqb2766)
Can you pls read the forum sticky about not posting just questions without any working/attempt/clear description of your problem. It's at the top of the forum.
https://www.thestudentroom.co.uk/sho....php?t=4919248

This is the 5th or 6th one today. Also, don't reply via PMs. Reply to the thread pls.
I said at the top i was gonna post working give it some time
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anya2003
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What paper is this from?
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nfr28
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Did you ever find the solution to this question?
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Nychus25
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Do you have the paper this is from?
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Arvind54
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Here is the answer for part d
Since the dentist requires 2 mins. to complete a routine appointment, the appropriate probability model will be truncated normal distribution defined T>2. then
P(T <= t given T>2) = P(T<= t, T>2) / P(T>2) =0.5 (median 50%)
P(2<T<= t) = 0.5 x P(T>2) = 0.5 x 0.1956829692 = 0.4021585154,
P(T<t) - P(T<2) = 0.4021585154,
P(T<t) = 0.4021585154 + P(T<2),
P(T<t) = 0.4021585154 +0.1956829692,
P(T<t) = 0.5978414846,
Use reverse normal to find t (Area=0.5978414846, sigma= sd= 3.5, mean= u=5, you will get XInv = 5.867173874, therefore t=5.87 (3sf)
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Asdfghjkl1000
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how do you do ciii
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