I need help with a z scores!!

Question

The probability that a single randomly selected bear in your sample would weigh between 200 and 300 pounds. with standard deviation of 107 and the mean of 206.

b)Are there any unusual weights in your sample? Provide evidence.

c)Using your sample data, are there any unusual weights in population?

d)Here are the population parameters: = 207.6 = 96.6 a) Using this data, find the percentage of bears that weigh between 100 and 300 pounds.

e)Find the weight of a bear for the 90th percentile

Answer 1

z statistic for a single bear = bear weight - mean / standard deviation.

To compute the probability that single bear is between 200 and 300 pounds, we first need to compute the z-statistics for 200 pounds:

z = 200-206 / 107 = -0.056

Then we need to compute the z-statistic for 300 pounds

z = 300-206/107 = 0.879

Now we need to use the normal probability tables or a calculator to find the probability (P) that the z statistic lies between those two values, i.e.,

P(-0.056 < z < 0.879) = P(z < 0.879) - P(z < -0.056) = 0.8106 - 0.4761 = 0.3345

would not be able to answer b) and c) without seeing sample data.

for part d would use similar procedure to find probability (or proportion) of bears between 100 and 300 pounds with population mean = 207.6 and population standard deviation = 96.6

Compute z stat for 100 pounds: (100 - 207.6) / 96.6 = -1.114

Compute z stat for 300 pounds: (300- 207.6) / 96.6 = 0.957

Then compute P(-1.114 < z < 0.957) = P(z < 0.957) - P(z < -1.114) = 0.8315 - 0.1335 = 0.698

To find the weight of the bear at the 90th percentile, we first need to find the z-value that corresponds to the 90th percentile, the z-value for which P(z < z-value) = 0.9. From the normal probability table that z-value is 1.282.

To find the bear weight that corresponds to that we use the z-statistic computation, knowing the z value is 1.282, the mean is 207.6 and the standard deviation is 96.6

1.282 = (bear weight - 207.6) / 96.6

bear weight = 96.6 * 1.282 + 207.6 = 331.44