Question

# 1. A particular fruit's weights are normally distributed, with a mean of 416 grams and a...

1. A particular fruit's weights are normally distributed, with a mean of 416 grams and a standard deviation of 11 grams.

The heaviest 17% of fruits weigh more than how many grams?

2. The patient recovery time from a particular surgical procedure is normally distributed with a mean of 4 days and a standard deviation of 1.9 days. Let X be the recovery time for a randomly selected patient. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(,)

b. What is the median recovery time?  days

c. What is the Z-score for a patient that took 6 days to recover?

d. What is the probability of spending more than 3.4 days in recovery?

e. What is the probability of spending between 3.7 and 4.4 days in recovery?

f. The 90th percentile for recovery times is  days.

1. Given mean = 416 and SD = 11

The heaviest 17% of fruits weigh more than how many grams is

Mean + Z0.83 * SD = 416 + 0.9542*11 = 426.4962 grams

since P(Z > 0.9542) = 0.17 from standard normal tables

2. a) X follows N(Mean = 4, SD = 1.9)

b) median = 4
since normal distribution is symmetric distribution i.e. mean = median = mode = 4

c) Z -score = (6 - Mean)/SD = (6-4)/1.9 =1.0526

d) P(X> 3.4 days) = P(z > (3.4 - Mean)/SD) = P(Z > (3.4 - 4)/1.9)) = P(Z > -0.31579) = 0.62392

e) P(3.7 < X < 4.4) = P( (3.7 - Mean)/SD < X < (4.4 - Mean)/SD)

= P( (3.7 - 4)/1.9 < X < (4.4 -4)/1.9)

= P( -0.1579 < X < 0.21053)

= P(Z < 0.21053) - P(Z < -0.1579)

= 0.58337 - 0.43727

= 0.1461

f) P(Z < 1.282) = 0.90

Recovery times is Mean + Z0.90 SD = 4 + 1.282*1.9 = 6.4358

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