A hospital specialises in heart surgery. During the year 2018-19, 2000 patients were admitted for treatment. Average payment made by a patient was Rs 120000 with a standard deviation of rs 25000.Assuming the distribution of payments to be normal find
1) The number of patients who paid between Rs 100000 and Rs 175000
2) The probability that the patients bill exceeds Rs 75000
3) The maximum amount paid by the lowest paying one third patient.
Answers
Given : A hospital specialises in heart surgery. During the year 2018-19, 2000 patients were admitted for treatment. Average payment made by a patient was Rs 120000 with a standard deviation of rs 25000.
To find : 1) The number of patients who paid between Rs 100000 and Rs 175000
2) The probability that the patients bill exceeds Rs 75000
3) The maximum amount paid by the lowest paying one third patient.
Solution:
Mean = 120000
Standard Deviation SD = 25000
Z Score = ( Value - Mean)/SD
1) The number of patients who paid between Rs 100000 and Rs 175000
Z score for 100000 = ( 100000 - 120000)/25000 = =0.8 => 0.2119 below this
Z score for 175000 = ( 175000 - 120000)/25000 = 2.2 => .9861 below this
Between Rs 100000 and Rs 175000 = 0.9861 - 0.2119 = 0.7742
2000 * 0.7742 = 1548.4 = 1548
1548 patients who paid between Rs 100000 and Rs 175000
patients bill exceeds Rs 75000
=> Z score for 75000 = ( 75000 - 120000)/25000 = -1.8 => 0.0359 below this
probability that the patients bill exceeds Rs 75000 = 1 - 0.0359 = 0.9641
The probability that the patients bill exceeds Rs 75000 = 0.9641
3) The maximum amount paid by the lowest paying one third patient.
lowest paying one third patient = 0.3333 => Z score = -0.435
-0.435 = ( Value - 120000)/25000
=> Value = 1,09,125
maximum amount paid by the lowest paying one third patient. = Rs 1,09,125
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