Suppose a normal distribution has a mean of 50 and a standard deviation of
5. What is the probability that a data value is between 48 and 52? Round your
answer to the nearest tenth of a percent.

Answer:

27.8% I think

Step-by-step explanation:

Then standardize the two scores

48 and 52

to be able to consult the table.

The formula for changing a raw score x into a z score is

Z=X−μσ.

For this problem, X is normal with

μ=50 and σ=5.

So, by symmetry of the standard normal curve,

P(48<X<52)

=P(48−50<X−50<52−50)

=P(48−505<X−505<52−505)

=P(−25<Z<25)

=P(−.40<Z<.40)

=P(Z<.40)−P(Z<−.40)

=P(Z<.40)−P(Z>.40)

=P(Z<.40)−(1−P(Z<.40))

=2P(Z<.40)−1

=2(.6554)−1

=1.3108−1

=.3108≈.31

=31%.

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