The reading speed of second grade students in a large city is approximately normal, with a mean of 90 words per minute (wpm) and a standard deviation of 10 wpm. complete parts (a) through (f). (a) what is the probability a randomly selected student in the city will read more than 96 words per minute?

Given : The reading speed of second grade students in a large city is approximately normal, with a mean of [tex]\mu=90[/tex] words per minute (wpm) and a standard deviation of [tex]\sigma=10[/tex] wpm.

Let x be the random variable that represents the reading speed of second grade students.

z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x= 96 words per minute

[tex]z=\dfrac{96-90}{10}=0.6[/tex]

Now, the probability a randomly selected student in the city will read more than 96 words per minute will be :-

[tex]P(x>96)=P(z>0.6)=1-P(\leq0.6)\\\\=1- 0.7257469=0.2742531\approx0.2743[/tex]

Hence, the probability a randomly selected student in the city will read more than 96 words per minute = 0.2743

maheshpatelvVT maheshpatelvVT · Answer 2 · 2022-04-13T09:39:20+02:00

The probability a randomly selected student in the city will read more than 96 words per minute is 0.2743.

It is given that the reading speed of second-grade students in a large city is approximately normal, with a mean of 90 words per minute (wpm) and a standard deviation of 10 wpm.

It is required to find the probability a randomly selected student in the city will read more than 96 words per minute.

What is a normal distribution?

It is defined as the continuous distribution probability curve which is most likely symmetric around the mean. At Z=0, the probability is 50-50% on the Z curve. It is also called a bell-shaped curve.

We know the formula for z-score:

[tex]\rm Z_{score}=\frac{x-\mu}{\sigma}[/tex]

We have,

x = 96 wpm

[tex]\rm \mu = 90[/tex] wpm

[tex]\rm \sigma = 10 \ wpm[/tex]

[tex]\rm Z_{score}=\frac{96-90}{10}[/tex]

[tex]\rm Z_{score}=0.6[/tex]

For the probability, a randomly selected student in the city will read more than 96 words per minute:

P(x>96) = p(z>0.6)

[tex]\rm 1- p(Z\leq 0.6)[/tex]

From the Z-table the value of [tex]\rm (Z\leq 0.6)[/tex] = 0.7257

[tex]=\rm 1- 0.7257[/tex]

= 0.2743

Thus, the probability a randomly selected student in the city will read more than 96 words per minute is 0.2743.

Know more about the normal distribution here:

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