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kow [346]
3 years ago
15

Assume the random variable x is normally distributed with mean u = 90 and standard deviation o=5. Find the indicated probability

.
P(X<80) =
(Round to four decimal places as needed.)
Mathematics
1 answer:
Setler79 [48]3 years ago
6 0

<u>Given</u>:

Let the random variable x is normally distributed with mean \mu=90 and \sigma=5

We need to determine the probability of P(X

<u>Probability of </u>P(X<u>:</u>

The formula to determine the value of P(X is given by

Z=\frac{X-\mu}{\sigma}

Thus, we have;

P(X

Simplifying, we get;

P(X

P(X

Using the normal distribution table, the value of -2 is given by 0.0228

P(X

Thus, the value of P(X is 0.0228

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Which is the interquartile range of the number of books read by the students in Mr. Ford's class?
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Over a nine-month period, OutdoorGearLab tested hardshell jackets designed for ice climbing, mountaineering, and backpacking. Ba
Yuri [45]

Answer:

(a) The mean is 65.9, the median is, 66.5 and the mode is. 61.

(b) The first and third quartiles are 61 and 71 respectively.

(c) The value of the 90th percentile is, 78.3.

Step-by-step explanation:

The ratings of 20 top-of-the line jackets, arranged in ascending order are as follows:

S = {42 , 53 , 54 , 61 , 61 , 61 , 62 , 63 , 64 , 66 , 67 , 67 , 68 , 69 , 71 , 71 , 76 , 78 , 81 , 83}

(a)

Compute the mean as follows:

\bar x=\frac{1}{n}\sum\limits^{n}_{i=1}{x_{i}}=\frac{1}{20}\times 1318=65.9

Compute the median as follows:

For an even number of values the median is the average of the middle two values.

M=\frac{10^{th}obs.+11^{th}obs.}{2}=\frac{66+67}{2}=66.5

Compute the mode as follows:

The mode of a dataset is the term that occurs most of the time.

m=61

Thus, the mean is 65.9, the median is, 66.5 and the mode is. 61.

(b)

The first quartile is the median of the first half of the data set.

S₁ = {42 , 53 , 54 , 61 , 61 , 61 , 62 , 63 , 64 , 66}

Q_{1}=\text{Median of }S_{1}=\frac{5^{th}obs.+6^{th}obs.}{2}=\frac{61+61}{2}=61

The third quartile is the median of the second half of the data set.

S₂ = {67 , 67 , 68 , 69 , 71 , 71 , 76 , 78 , 81 , 83}

Q_{3}=\text{Median of }S_{2}=\frac{15^{th}obs.+16^{th}obs.}{2}=\frac{71+71}{2}=71

Thus, the first and third quartiles are 61 and 71 respectively.

(c)

The <em>p</em>th percentile is a data value such that at least p% of the data set is less-than or equal to this data value and at least (100 - p)% of the data-set are more-than or equal to this data value.

Use the Excel formula "=PERCENTILE.INC(array,0.9)" to compute the 90th percentile.

The value of the 90th percentile is, 78.3.

This value implies that 90% of the ratings are less than 78.3.

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