Fill in the blank: 50% of the area under the standard normal curve is between -0.5 and

@phi

@hartnn

@Directrix

About 50% of the area under a normal curve falls within ±0.5 standard deviations of the mean.

Well it's more like ±0.67, but with the way the question was worded... http://blog.theoptionstradingcourse.com/wp-content/uploads/2011/09/normal671.gif

50% of the area under the standard normal curve is between z = -0.5 and z = 0.874

thanks @kropot72

You're welcome :)

@kropot72 The 60th percentile of a normal distribution is 110. The 70th percentile is 120. Find the 90th percentile. please help

please help @kropot72

please answer @Directrix

The 60th percentile of a normal distribution is 110. The 70th percentile is 120. Find the 90th percentile. please help

I'll add my work shortly. I got 148.15 as the 90th percentile. Not sure that is correct.

I thought we were finished. I still have to enter that other work that supports the answer.

The 60th percentile is z = 0.253\[z _{60th}=0.253=\frac{110-\mu}{\sigma}\] The 70th percentile is z = 0.524 \[z _{70th}=0.534=\frac{120-\mu}{\sigma}\]

The 90th percentile is z = 1.282 After solving for mu and sigma and substituting I got X = 147.97 for the 90th percentile,

please help

@Nurali Post this problem in a new thread. If @kropot72 and I do all this work, we expect to be "paid." :)

^^

Urgent @Nurali --> Move the question to a new thread so that we can get to work.

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